Developments in Mathematics

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VOLUME 24

Series Editors: Krishnaswami Alladi, University of Florida Hershel M. Farkas, Hebrew University of Jerusalem Robert Guralnick, University of Southern California

For further volumes: www.springer.com/series/5834 Jerzy Kakol ˛ Wiesław Kubis´ Manuel López-Pellicer

Descriptive Topology in Selected Topics of Functional Analysis Jerzy Kakol ˛ Wiesław Kubis´ Faculty of Mathematics and Informatics Institute of Mathematics A. Mickiewicz University Jan Kochanowski University 61-614 Poznan 25-406 Kielce Poland Poland [email protected] and Institute of Mathematics Manuel López-Pellicer Academy of Sciences of the Czech Republic IUMPA 115 67 Praha 1 Universitat Poltècnica de València Czech Republic 46022 Valencia [email protected] Spain and Royal Academy of Sciences 28004 Madrid Spain [email protected]

ISSN 1389-2177 ISBN 978-1-4614-0528-3 e-ISBN 978-1-4614-0529-0 DOI 10.1007/978-1-4614-0529-0 Springer New York Dordrecht Heidelberg London

Library of Congress Control Number: 2011936698

Mathematics Subject Classiﬁcation (2010): 46-02, 54-02

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Springer is part of Springer Science+Business Media (www.springer.com) To our Friend and Teacher Prof. Dr. Manuel Valdivia Preface

We invoke (descriptive) topology recently applied to (functional) analysis of infinite-dimensional topological vector spaces, including Fréchet spaces, (LF)- spaces and their duals, Banach spaces C(X) over compact spaces X, and spaces Cp(X), Cc(X) of continuous real-valued functions on a completely regular Haus- dorff space X endowed with pointwise and compact–open topologies, respectively. The (LF)-spaces and duals particularly appear in many fields of functional analysis and its applications: distribution theory, differential equations and complex analysis, to name a few. Our material, much of it in book form for the first time, carries forward the rich legacy of Köthe’s Topologische lineare Räume (1960), Jarchow’s Locally Convex Spaces (1981), Valdivia’s Topics in Locally Convex Spaces (1982), and Pérez Car- reras and Bonet’s Barrelled Locally Convex Spaces (1987). We assume their (standard English) terminology. A topological vector space (tvs) must be Hausdorff and have a real or complex scalar field. A locally convex space (lcs) is a tvs that is locally convex. Engelking’s General Topology (1989) serves as a default reference for general topology. The authors wish to thank Professor B. Cascales, Professor M. Fabian, Professor V. Montesinos, and Professor S. Saxon for their valuable comments and suggestions, which made this material much more readable. The research of J. Kakol ˛ was partially supported by the Ministry of Science and Higher Education, Poland, under grant no. NN201 2740 33. W. Kubis´ was supported in part by grant IAA 100 190 901, by the Institutional Research Plan of the Academy of Sciences of the Czech Republic under grant no. AVOZ 101 905 03, and by an internal research grant from Jan Kochanowski Uni- versity in Kielce, Poland. The research of J. Kakol ˛ and M. López-Pellicer was partially supported by the Spanish Ministry of Science and Innovation, under project no. MTM 2008-01502. Poznan, Poland Jerzy Kakol ˛ Kielce, Poland Wiesław Kubis´ Valencia, Spain Manuel López-Pellicer

vii Contents

1 Overview ...... 1 1.1 General comments and historical facts ...... 7 2 Elementary Facts about Baire and Baire-Type Spaces ...... 13 2.1 Baire spaces and Polish spaces ...... 13 2.2 A characterization of Baire topological vector spaces ...... 18 2.3 Arias de Reyna–Valdivia–Saxon theorem ...... 20 2.4 Locally convex spaces with some Baire-type conditions ...... 24 2.5 Strongly realcompact spaces X and spaces Cc(X) ...... 36 2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X) .. 46 2.7 Sequential conditions for locally convex Baire-type spaces ..... 56 3 K-analytic and Quasi-Suslin Spaces ...... 63 3.1Elementaryfacts...... 63 3.2 Resolutions and K-analyticity ...... 71 3.3 Quasi-(LB)-spaces ...... 82 3.4 Suslin schemes ...... 91 3.5 Applications of Suslin schemes to separable metrizable spaces . . . 93 3.6 Calbrix–Hurewicz theorem ...... 101 4 Web-Compact Spaces and Angelic Theorems ...... 109 4.1 Angelic lemma and angelicity ...... 109 4.2 Orihuela’s angelic theorem ...... 111 4.3 Web-compact spaces ...... 113 4.4 Subspaces of web-compact spaces ...... 116 4.5 Angelic duals of spaces C(X) ...... 118 4.6 About compactness via distances to function spaces C(K) .....120 5 Strongly Web-Compact Spaces and a Closed Graph Theorem ....137 5.1 Strongly web-compact spaces ...... 137 5.2 Products of strongly web-compact spaces ...... 138 5.3 A closed graph theorem for strongly web-compact spaces .....140

ix x Contents

6 Weakly Analytic Spaces ...... 143 6.1 A few facts about analytic spaces ...... 143 6.2 Christensen’s theorem ...... 149 6.3 Subspaces of analytic spaces ...... 155 6.4 Trans-separable topological spaces ...... 157 6.5 Weakly analytic spaces need not be analytic ...... 164 6.6 More about analytic locally convex spaces ...... 167 6.7 Weakly compact density condition ...... 168 6.8 More examples of nonseparable weakly analytic tvs ...... 174 7 K-analytic Baire Spaces ...... 183 7.1 Baire tvs with a bounded resolution ...... 183 7.2 Continuous maps on spaces with resolutions ...... 187 8 A Three-Space Property for Analytic Spaces ...... 193 8.1AnexampleofCorson...... 193 8.2 A positive result and a counterexample ...... 196

9 K-analytic and Analytic Spaces Cp(X) ...... 201 9.1 A theorem of Talagrand for spaces Cp(X) ...... 201 9.2 Theorems of Christensen and Calbrix for Cp(X) ...... 204 9.3 Bounded resolutions for Cp(X) ...... 215 9.4 Some examples of K-analytic spaces Cp(X) and Cp(X, E) ....230 9.5 K-analytic spaces Cp(X) over a locally compact group X .....231 ∧ 9.6 K-analytic group Xp ofhomomorphisms...... 234 10 Precompact Sets in (LM)-Spaces and Dual Metric Spaces ...... 239 10.1 The case of (LM)-spaces: elementary approach ...... 239 10.2 The case of dual metric spaces: elementary approach ...... 241 11 Metrizability of Compact Sets in the Class G ...... 243 11.1 The class G:examples...... 243 11.2 Cascales–Orihuela theorem and applications ...... 245 12 Weakly Realcompact Locally Convex Spaces ...... 251 12.1 Tightness and quasi-Suslin weak duals ...... 251 12.2 A Kaplansky-type theorem about tightness ...... 254 12.3 K-analytic spaces in the class G ...... 258 12.4 Every WCG Fréchet space is weakly K-analytic ...... 260 12.5 Amir–Lindenstrauss theorem ...... 266 12.6AnexampleofPol...... 271 12.7 More about Banach spaces C(X) over compact scattered X ....276 13 Corson’s Property (C) and Tightness ...... 279 13.1 Property (C) and weakly Lindelöf Banach spaces ...... 279 13.2 The property (C) for Banach spaces C(X) ...... 284 Contents xi

14 Fréchet–Urysohn Spaces and Groups ...... 289 14.1 Fréchet–Urysohn topological spaces ...... 289 14.2 A few facts about Fréchet–Urysohn topological groups ...... 291 14.3 Sequentially complete Fréchet–Urysohn spaces are Baire .....296 14.4 Three-space property for Fréchet–Urysohn spaces ...... 299 14.5 Topological vector spaces with bounded tightness ...... 302 15 Sequential Properties in the Class G ...... 305 15.1 Fréchet–Urysohn spaces are metrizable in the class G ...... 305 15.2 Sequential (LM)-spaces and the dual metric spaces ...... 311 − 15.3 (LF )-spaces with the property C3 ...... 320 16 Tightness and Distinguished Fréchet Spaces ...... 327 16.1 A characterization of distinguished spaces ...... 327 16.2 G-basesandtightness...... 334 16.3 G-bases, bounding, dominating cardinals, and tightness .....338 16.4 More about the Wulbert–Morris space Cc(ω1) ...... 349 17 Banach Spaces with Many Projections ...... 355 17.1 Preliminaries, model-theoretic tools ...... 355 17.2 Projections from elementary submodels ...... 361 17.3 Lindelöf property of weak topologies ...... 364 17.4 Separable complementation property ...... 365 17.5 Projectional skeletons ...... 369 17.6 Norming subspaces induced by a projectional skeleton ...... 375 17.7 Sigma-products ...... 380 17.8 Markushevich bases, Plichko spaces and Plichko pairs ...... 383 17.9 Preservation of Plichko spaces ...... 388 18 Spaces of Continuous Functions over Compact Lines ...... 395 18.1 General facts ...... 395 18.2 Nakhmanson’s theorem ...... 398 18.3 Separable complementation ...... 399 19 Compact Spaces Generated by Retractions ...... 405 19.1Retractiveinversesystems...... 405 19.2 Monolithic sets ...... 409 19.3 Classes R and RC ...... 411 19.4 Stability ...... 412 19.5Someexamples...... 415 19.6 The ﬁrst cohomology functor ...... 418 19.7 Compact lines ...... 422 19.8 Valdivia and Corson compact spaces ...... 425 19.9 Preservation theorem ...... 432 19.10 Retractional skeletons ...... 434 19.11 Primarily Lindelöf spaces ...... 438 19.12 Corson compact spaces and WLD spaces ...... 440 19.13 A dichotomy ...... 442 xii Contents

19.14 Alexandrov duplications ...... 446 19.15 Valdivia compact groups ...... 448 19.16 Compact lines in class R ...... 451 19.17 More on Eberlein compact spaces ...... 456 20 Complementably Universal Banach Spaces ...... 467 20.1Amalgamationlemma...... 467 20.2 Embedding-projection pairs ...... 469 20.3 A complementably universal Banach space ...... 471 References ...... 475 Index ...... 491 Chapter 1 Overview

Let us briefly describe the organization of the book. Chapter 2, essential to the sequel, contains classical results about Baire-type conditions (Baire-like, b-Baire-like, CS-barrelled, s-barrelled) on tvs. We include applications to closed graph theorems and C(X) spaces. We also provide the first proof in book form of a remarkable result of Saxon [355] (extending earlier results of Arias de Reyna and Valdivia) that states that, under Martin’s axiom, every lcs containing a dense hyperplane contains a dense non-Baire hyperplane. Ours, then, is the first book to solve the first problem formally posed in Pérez Carreras and Bonet’s excellent monograph. Chapter 2 also contains analytic characterizations of certain completely regular Hausdorff spaces X. For example, we show that X is pseudocompact, is Warner bounded, or Cc(X) is a (df )-space if and only if for each sequence (μn)n in the dual Cc(X) there exists a sequence (tn)n ⊂ (0, 1] such that (tnμn)n is weakly bounded, strongly bounded, or equicontinuous, respectively ([231, 232]). These characterizations help us produce a (df )-space Cc(X) that is not a (DF )- space [232], solving a basic and long-standing open question. The third characterization is joined by nine more that supply tenfold an implied Jarchow request. These forge a strong link we happily claim between his book and ours. Chapter 3 deals with the K-analyticity of a topological space E and the con- { : ∈ NN} cept of a resolution generated on E (i.e., a family of sets Kα α such = ⊂ ≤ that E α Kα and Kα Kβ if α β). Compact resolutions (i.e., resolutions N {Kα : α ∈ N } whose members are compact sets) naturally appear in many situations in topology and functional analysis. Any K-analytic space admits a compact resolution [388], and for many topological spaces X the existence of such a resolution is enough for X to be K-analytic; (see [80], [82]). Many of the ideas in the book are related to the concept of compact resolution and are already in or have been inspired by papers [388], [80], [82]. It is an easy and elementary exercise to observe that any separable metric and complete space E admits a compact resolution, even swallowing compact sets. In Chapter 3, we gather some results, mostly due to Val- divia [421], about lcs (called quasi-(LB)-spaces) admitting resolutions consisting of Banach discs and their relations with the closed graph theorems.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 1 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_1, © Springer Science+Business Media, LLC 2011 2 1Overview

These concepts are related to another one, called a Suslin scheme, which provides a powerful tool to study structural properties of metric separable spaces (see [175], [346]). Chapter 3 presents Hurewicz and Alexandrov’s theorems as well as the Calbrix–Hurewicz theorem, which yields that a regular analytic space X (i.e., a continuous image of the space NN)isnotσ -compact if and only if X contains a closed subset homeomorphic to NN. We have tried to present proofs in a transparent form. The reader is also referred to the magnificent works [387], [388], [421], [346], [99], among others. Chapter 4 deals with the class of angelic spaces, introduced by Fremlin, for which several variants of compactness coincide. A remarkable paper of Orihuela [320]introduces a large class of topological spaces X (under the name web-compact)for which the space Cp(X) is angelic. Orihuela’s theorem covers many already known partial results providing Eberlein–Šmulian-type results. Following Orihuela [320], we show that Cp(X) is angelic if X is web-compact. This yields, in particular, Ta- lagrand’s result [388] stating that for a compact space X the space Cp(X) is K- analytic if and only if C(X) is weakly K-analytic. In Chapter 4, we present some quantitative versions of Grothendieck’s characterization of the weak compactness for spaces C(X) (for compact Hausdorff spaces X) and quantitative versions of the classical Eberlein–Grothendieck and Krein–Šmulian theorems. We follow very recent works of Angosto and Cascales [6], [7], [10], Angosto [9], Angosto, Cascales and Namioka [8], Cascales, Marciszewski and Raja [92], Hájek, Montesinos and Zizler [150] and Granero [187]. The last two articles, [150] and [187], where in the case of Banach spaces these quantitative generalizations have been studied and presented, motivated the other papers mentioned. In Chapter 5, we continue the study of web-compact spaces. A subclass of web- compact spaces, called strongly web-compact, is introduced, and a closed graph theorem for such spaces is provided. We prove that an own product of a strongly web-compact space need not be web-compact. This shows that there exists a quasi- Suslin space X such that X × X is not quasi-Suslin. Chapter 6 studies analytic spaces. We show that a regular space X is analytic if and only if X has a compact resolution and admits a weaker metric topology. This fact, essentially due to Talagrand [392], extended Choquet’s theorem [97](every metric K-analytic space is analytic); see also [85]. Several applications will be provided. We show Christensen’s theorem [99] stating that a separable metric topological space X is a Polish space if and only if X admits a compact resolution swallowing compact sets. The concept of a compact resolution swallowing compact sets is already present in the main result of [82, Theorem 1]. We study trans-separable spaces and show that a tvs with a resolution of precompact sets is trans-separable [342]. This serves to prove [82] that precompact sets are metrizable in any uniform space whose uniformity admits a U -basis. Consequences are provided. Chapter 6 also works with the following general problem (among some others): When can analyticity or K-analyticity of the weak topology σ(E,E) of a dual pair (E, E) be lifted to stronger topologies on E compatible with the dual pair? The question is essential since (as we show) there exist many weakly analytic lcs’s (i.e., analytic in the weak topology σ(E,E) that are not analytic. We prove that if X is 1Overview 3 an uncountable analytic space, the Mackey dual Lμ(X) of Cp(X) is weakly analytic and not analytic. The density condition due to Heinrich [203], studied in a series of papers of Bierstedt and Bonet [51], [52], [53], [54], [55], motivates a part of Chapter 6 that studies the analyticity of the Mackey and strong duals of (LF )-spaces. In Chapter 7, we show that a tvs that is a Baire space and admits a countably compact resolution is metrizable, separable and complete. This extends a classical result of De Wilde and Sunyach [111] and Valdivia’s theorem [421]. An interesting recent applicable result due to Drewnowski (highly inspired by [233]) about continuous maps between F-spaces is presented. Namely, we show that a linear map N T : E → F from an F-space E having a resolution {Kα : α ∈ N } into a tvs F is continuous if each restriction T |Kα is continuous. This theorem was motivated by the Arias–De Reina–Valdivia–Saxon theorem about non-Baire dense hyperplanes in Banach spaces. We provide a large class of weakly analytic metrizable and separable Baire tvs’s not analytic (clearly such spaces are necessarily not locally convex). Examples of spaces of this type will be used in Chapter 8 to prove that analyticity is not a three-space property. We prove, however, that a metrizable topological vector space E is analytic if it contains a complete locally convex analytic subspace F such that the quotient E/F is analytic. We reprove also in Chapter 8 (using Corson’s example [103]) that the Lindelöf property is not a three-space property. Chapter 9 partially continues the study started in Chapter 3 and deals with K- analytic and analytic spaces Cp(X). Some results due to Talagrand [388], Tkachuk [399], Velichko [27] and Canela [78] are presented. We extend the main result of [399] characterizing K-analytic spaces Cp(X) in terms of resolutions. Christensen’s remarkable theorem [99] stating that a metrizable and separable space X is σ - compact if and only if Cp(X) is analytic is proved. We show that the analyticity of Cp(X) for any X implies that X is σ -compact, from Calbrix [77]. A characterization of σ -compactness of a cosmic space X in terms of subspaces of RX is provided from Arkhangel’skii and Calbrix [30]. Finally, we show that Cp(X) is K- X analytic-framed in R if and only if Cp(X) admits a bounded resolution [160]. We also collect several equivalent conditions for a space Cp(X) to be a Lindelöf space over locally compact groups X;see[234]. Chapter 10, which might be a good motivation for Chapters 11 and 12, extends the main result of Cascales and Orihuela [81] and presents the unified and direct proofs [229]ofPfister[329], Cascales and Orihuela [81] and Valdivia’s [423] theorems about the metrizability of precompact sets in (LF )-spaces, (DF )-spaces and dual metric spaces, respectively. The proofs from [229] do not require the typical machinery of quasi-Suslin spaces, upper semicontinuous compact-valued maps, and so on. Chapter 11 introduces (after Cascales and Orihuela [83]) a large class of locally convex spaces under the name the class G.AnlcsE is said to be in the class G if its N topological dual E admits a resolution {Aα : α ∈ N } such that sequences in each Aα are equicontinuous. The class G contains among others all (LM)-spaces (hence (LF )-spaces) and dual metric spaces (hence (DF )-spaces), spaces of distributions D(Ω), spaces A(Ω) of real analytic functions on open Ω ⊂ Rn and others. We 4 1Overview show in Chapter 11 the main result of [83], with a simpler proof from [155], stating that every precompact set in an lcs in the class G is metrizable. This general result covers many already known theorems for (DF )-spaces, (LF )-spaces and dual metric spaces. In Chapter 12, we continue the study of spaces in the class G. We prove that the weak∗ dual (E,σ(E,E)) of an lcs E in the class G is K-analytic if and only if (E,σ(E,E)) is Lindelöf if and only if (E, σ (E, E)) has countable tightness if and only if each finite product (E,σ(E,E))n is Lindelöf; see [88]. Develop- ing the argument producing upper semicontinuous maps, we also show that every quasibarrelled space in the class G has countable tightness both for the weak and the original topologies. This extends a classical result of Kaplansky for a metrizable lcs; see [165]. Although (DF )-spaces belong to the class G, concrete examples of (DF )-spaces without countable tightness are provided. On the other hand, there are many Banach spaces E for which E endowed with the weak topology σ(E,E) is not Lindelöf. We show, however, from Khurana [242], that every weakly compactly generated (WCG) Fréchet space E is weakly K-analytic (i.e., (E, σ (E, E)) is K-analytic). This extends Talagrand’s corresponding result for WCG Banach spaces; see [390] and also [149], [323]. In general, for a WCG lcs, this result fails, as Hunter and Lloyd [209] have shown. It is natural to ask (Corson [103];seealso[271]) if every weakly Lindelöf Banach space is a WCG Banach space (i.e., if E admits a weakly compact set whose linear span is dense in E). The first example of a non-WCG Banach space whose weak topology is Lindelöf was provided by Rosenthal [349]. We present an example due to Pol [336]showing that there exists a Banach space C(X) over a compact scattered space X such that C(X) is weakly Lindelöf and C(X) is not a WCG Banach space. This example also answers (in the negative) some questions of Corson [103, Problem 7], posed by Benyamini, Rudin and Wage from [49]. Talagrand, inspired and motivated by several results of Corson, Lindenstrauss and Amir, continued this line of research in his remarkable papers (see, e.g., [388], [389], [390], [391]). Chapter 12 also contains the proof of the Amir–Lindenstrauss theorem that every nonseparable reflexive Banach space contains a complemented separable subspace [270]. Several consequences are provided. This subject, related to WCG Banach spaces and the Amir–Lindenstrauss theorem, will be continued in Chapters 17, 18, 19 and 20, where Banach spaces with a rich family of projections onto separable subspaces are studied. In Chapter 13, the class of Banach spaces having the property (C) is studied. This property, isolated by Corson [103], provides a large subclass of Banach spaces E whose weak topology need not be Lindelöf. We collect some results of Corson [103], Pol [334], [338] and Frankiewicz, Plebanek and Ryll-Nardzewski [168]. Chapters 14 and 15 deal with topological (vector) spaces satisfying some sequential conditions. We study Fréchet–Urysohn spaces (i.e., spaces E such that for each A ⊂ E and each x ∈ A there exists a sequence in A converging to x). The main result states that every sequentially complete Fréchet–Urysohn lcs is a Baire space. Since every infinite-dimensional Montel (DF )-space E is nonmetrizable and sequential (i.e., every sequentially closed set in E is closed), the following question arises: Is every Fréchet–Urysohn space in the class G metrizable? 1Overview 5

In Chapter 15, we prove that an lcs in the class G is metrizable if and only if E is b-Baire-like and if and only if E is Fréchet–Urysohn; see [88], [89]. Consequently, no proper (LB)-space E is Fréchet–Urysohn (since E contains the space ϕ; i.e., the ℵ0-dimensional vector space with the finest locally convex topology). We prove that if a (DF )-or(LM)-space E is sequential, then E is either metrizable or Montel (DF );see[229]. Webb [415] introduced the property C3 (i.e., sequential closure of any set is sequentially closed), which characterizes metrizability for (LM)-spaces but not for (DF )-spaces. We distinguish a variant of the property C3 called prop- − erty C3 (i.e., sequential closure of any vector subspace is sequentially closed) and − characterize both (DF )-spaces and (LF )-spaces with the property C3 as being of the form M, φ,orM × φ, where M is metrizable [229]. In Chapter 16, we apply the concept of tightness to study distinguished Fréchet spaces. Valdivia provided a nondistinguished Fréchet space whose weak∗ bidual is quasi-Suslin but not K-analytic; see [421]. Using the concept of tightness, we show that Köthe’s echelon nondistinguished Fréchet space λ1(A) serves the same purpose [157], and we provide another (much simpler) proof of the deep result of Bastin and Bonet stating that for λ1(A) there exists a locally bounded discontinuous linear functional over the space (λ1,β(λ1,λ1));see[158]. The basic fact [157] is that a (DF )-space is quasibarrelled if and only if its tightness is countable. We show that a Fréchet space is distinguished if and only if its strong dual has countable tightness. This approach to studying distinguished Fréchet spaces leads to a rich supply of (DF )-spaces whose weak∗ duals are quasi-Suslin but not K-analytic, for example spaces Cc(κ) for κ a cardinal of uncountable cofinality. The small cardinals b and d will be used to improve the analysis of Köthe’s example; see [159], [157]. The bounding cardinal b (introduced by Rothberger) is the smallest infinite- dimensionality for metrizable barrelled spaces; see [357] for details. In general, a quasibarrelled space E belongs to the class G if and only if E N admits a G-basis (i.e., a family {Uα : α ∈ N } of neighborhoods of zero in E such that every neighborhood of zero in E contains some Uα;see[157]). This concept provides spaces Cc(X) different from those Talagrand presented in [388], whose weak∗ dual is not K-analytic but does have a compact resolution. We show that the weak∗ dual of any space in the class G is quasi-Suslin [159]. An immediate consequence is that every space with a G-basis enjoys this property. In Chapter 16, we show that Cc(ω1) may or may not have a G-basis. The existence of a G-basis for Cc(ω1) depends on the axioms of set theory. Cc(ω1) has a G-basis if and only if ℵ1 = b. Several interesting examples of (DF )-spaces that admit and do not admit G-bases will also be provided. In Chapter 17, we continue the subject developed in Chapter 12 related to WCG Banach spaces and the Amir–Lindenstrauss theorem. We discuss Banach spaces that have a rich family of norm-one projections onto separable subspaces. Probably, the most general class of Banach spaces with “many” projections specifies the separable complementation property. Recall that a Banach space E has the separable complementation property (SCP) if for every separable subspace D of E there exists a bounded linear projection with a separable range containing D. This property seems to be too general for proving any reasonable structural properties of Banach 6 1Overview spaces. A strengthening of the SCP is the notion of a projectional skeleton, defined and studied in Section 17.5. Banach spaces with a projectional skeleton have good stability properties as well as some nice structural ones. For instance, they have an equivalent locally uniformly convex norm and admit a bounded injective linear operator into some c0(Γ )-space. A natural property of a projectional skeleton is commutativity. It turns out that this is equivalent to the existence of a countably norming Markushevich basis. A space with this property is often called a Plichko space. We study this class of spaces in Sect. 17.7. A more special property of a projectional skeleton gives the class of WLD Banach spaces. We present selected results concerning the complementation property in general Banach spaces in order to motivate the study of projectional skeletons and projectional resolutions. This section contains some information about Plichko spaces, stability of this class and some natural examples. In order to simplify several arguments and constructions of projections, we present in Section 17.1 the method of elementary substructures coming from logic. Using this method we prove, for example, a result on the Lindelöf property of the topology induced by a certain norming subspace of the dual and by the projectional skeleton. In the case of WLD spaces, this is the well-known result on the Lindelöf property of the weak topology. The method of elementary substructures is also used in the following chapters for proving topological properties such as countable tightness. Chapter 18 discusses selected properties of Banach spaces of type C(X), where X is a linearly ordered compact space, called a compact line for short. In particular, we present Nakhmanson’s theorem stating that if X is a compact line such that Cp(X) is a Lindelöf space, then X is second-countable. Compact lines are relatively easy to investigate, yet they form a rich class of spaces and provide several interesting examples. A very special case is the smallest uncountable well-ordered space, ω1 +1, which appears several times in the previous chapters. Its space of continuous functions turns out to be a canonical example for several topological and geometric properties of Banach spaces. More complicated compact lines provide examples related to Plichko spaces. Chapter 19 presents several classes of nonmetrizable compact spaces that correspond to well-known classes of Banach spaces with many projections. In particular, we discuss the class of Valdivia compact spaces and its subclasses: Corson and Eberlein compact spaces. We discuss a general class of compact spaces obtained by limits of continuous retractive sequences. We also introduce the notion of a retractional skeleton, dual to projectional skeletons in Banach spaces. The last section of Chapter 19 contains an overview of Eberlein compact spaces, with some classical results and examples relevant to the subject of previous chapters. Finally, Chapter 20 deals with complementably universal Banach spaces. As- suming the continuum hypothesis, there exists a complementably universal Banach space of density ℵ1 for the class of Banach spaces with a projectional resolution of the identity. Similar methods produce a universal preimage for the class of Valdivia compact spaces of weight ℵ1. 1.1 General comments and historical facts 7

1.1 General comments and historical facts

The earliest approach to K-analytic spaces seems to be due to Choquet [96], who called a topological space K-analytic if it is a Kσδ-set in some compact space. Rogers proved [345] that if X is a completely regular Hausdorff space, the last condition is equivalent to the following one: (*) There exists a Polish space Y and an upper semicontinuous compact-valued map from Y covering X. The property (*) can be seen in Martineau [282]; see also Frolik [177], Rogers [345], Stegall [387] and Sion [380], [381]. In our book, by a K-analytic space X we mean a topological space satisfying condition (*). Since every Polish space is a continuous image of the space NN, we note another equivalent way to look at K-analytic spaces is as the image under an upper semicontinuous compact-valued map of the space NN. The images under upper semicontinuous compact set-valued maps of Polish spaces are also known in the literature as K-Suslin spaces; see also [421], [282], [177], [345]. K-analytic and analytic spaces are also useful topological objects for the study of nice properties of topological measures; see [371], [170]. For example, every semifinite topological measure that is inner regular for closed sets for a K-analytic space is inner regular for compact sets, and every semifinite Borel measure is inner regular for compact sets for an analytic space. Every K-analytic N space X admits a compact resolution (i.e., a family {Kα : α ∈ N } of compact sets covering X such that Kα ⊂ Kβ if α ≤ β); see [388], [80], [82], [131]—(in [131] this term was used formally for the first time). In the frame of angelic spaces X,the existence of a compact resolution implies that X is necessarily a K-analytic space [80]. Talagrand [388] had already observed this for spaces Cp(X) over compact spaces X; see also [85]. Many interesting topological problems in infinite-dimensional topological vector spaces might be motivated by some results from the theory of Cp(X) spaces. Let us mention, for example, one of them due to Velichko [27, Theorem I.2.1]: The space Cp(X) is σ -compact (i.e., covered by a sequence of compact sets) if and only if X is finite. Tkachuk and Shakhmatov [403] extended this result to σ -countably compact spaces Cp(X); see also [25] for a general approach including both cases. Clearly, if (Kn)n is an increasing sequence of compact sets covering Cp(X), then the sets := = ∈ NN Kα Kn1 , where α (nk) , form a compact resolution for Cp(X).Onthe other hand, we can prove that Cp(X) has a fundamental sequence of bounded sets only if X is finite, see Chapter 2. Natural questions arise: Characterize completely regular Hausdorff spaces X for which Cp(X) admits a compact resolution. When does Cp(X) admit a resolution consisting of topologically bounded sets? Recently, a related problem has been solved by Tkachuk [399], who proved: Cp(X) is K-analytic if and only if Cp(X) admits a compact resolution. We provide another approach to solving this problem. We show that if Cp(X) admits a resolution consisting of tvs-bounded sets (i.e., sets absorbed by any neigh- 8 1Overview borhood of zero in Cp(X)) then Cp(X) is angelic. Since angelic spaces with compact resolutions are K-analytic [80], this yields Tkachuk’s result and provides more applications. The class of weakly Lindelöf determined Banach spaces (WLD Banach spaces, introduced in [13]) provides a larger class of weakly Lindelöf Banach spaces containing the weakly compactly generated (WCG) Banach spaces. The study of WLD Banach spaces was motivated by results of Gul’ko [196] about weakly K-countably determined Banach spaces (also called weakly countably determined WCD Banach spaces, weakly Lindelöf Σ-spaces or Vašák spaces ([388], [418];seealso[147], [298]). Recall that according to [322] a Banach space E is WLD if and only if its closed unit ball in E is Corson compact in σ(E,E). Quite recently, Cascales, Namioka and Orihuela [91] have shown that if E is a Banach space satisfying Cor- son’s property (C) and E admits a projectional generator, then E is WLD. We reprove this result in Section 17.6 using the notion of a projectional skeleton. See also [84], [149] (and references therein) for more details. In 1961, H. Corson [103] started a systematic study of certain topological properties of the weak topology of Banach spaces. This line of research provided more general classes such as reflexive Banach spaces, weakly compactly generated Ba- nach spaces [5], [349], [121] and the class of weakly K-analytic and weakly K- countably determined Banach spaces. For another approach to studying geometric and topological properties of nonseparable Banach spaces, we refer to [200]. In his fundamental paper [103], Corson asked if WCG Banach spaces are exactly those Banach spaces whose weak topology is Lindelöf. The first example of a non- WCG Banach space whose weak topology was Lindelöf was provided by Rosenthal [349]. One can ask for which compact spaces X the space Cp(X) is Lindelöf. This problem was first studied by Corson [103];seealso[105]. The class of Corson compact spaces X (i.e., homeomorphic to a compact subset of a Σ-product of real lines) provides examples of Lindelöf spaces Cp(X);see[4], [195]. On the other hand, for every weakly K-analytic Banach space E, the closed unit ball in E in the topology σ(E,E) is a Corson compact set [196]. There exist, however, examples due to Talagrand, Haydon and Kunen (under the continuum hypothesis (CH)) of Corson compact spaces X such that the Banach space C(X) is not weakly Lindelöf; see [312]. Also, in [15] it was shown that for a Corson compact space X the Banach space C(X) is WLD if and only if every positive regular Borel measure on X has separable support. In general, the claim that for every Corson compact space X the space C(X) is a WLD Banach space is independent of the usual axioms of set theory [15]. There exist concrete Banach spaces C(X) over compact scattered spaces X that are weakly Lindelöf but not WCG; see, for example, [338], [388]. Nevertheless, for a compact space X, the Banach space C(X) is WCG if and only if X is Eberlein compact [5]; see also [147], [149]. For a compact space X, the space C(X) is weakly K-analytic if and only if Cp(X) is K-analytic [388]. This distinguishes the class of Talagrand compact spaces (i.e., compact X for which Cp(X) is K-analytic). This line of research between topology and functional analysis inspired several specialists (mainly from functional analysis) to develop new techniques from de- 1.1 General comments and historical facts 9 scriptive topology to study concrete problems and classes of spaces in linear functional analysis; see [149] for many references. For example, one may ask: (i) Is a Banach space E weakly Lindelöf if its weak∗ dual (E,σ(E,E)) has countable tightness? (ii) If σ(E,E) is Lindelöf, is the unit ball in E of countable tightness in σ(E,E)? Question (ii) is related to Banach spaces satisfying the property (C) of Corson. This property, introduced by Corson [103], provided a large subclass of Banach spaces E whose weak topology need not be Lindelöf. Papers of Corson and Pol described the property (C) in terms of countable tightness-type conditions for the topology σ(E,E). Corson’s paper [103] concerning the property (C) and results of Pol from [334], [336]or[337] motivated several articles, for example [323], [331], [332], [168], [88], among others, to study concrete classes of the weakly Lindelöf Banach spaces. This subject of research has been continued by many specialists; we refer the reader to articles [80], [83], [78], [293][294];seealso[88], [89], [85], [159], [157], [158], [131]. Many important spaces in functional analysis are defined as certain (DF )- spaces, (LB)-spaces, or (LF )-spaces (i.e., inductive limits of a sequence of Banach (Fréchet) spaces), or their strong duals; see, for example, [51], [52], [54], [288], [328], [213], [421] as good sources of information. A significant difference from the Banach space case is that the strong dual of a Fréchet space is not metrizable in general. The strong duals of Fréchet spaces are (DF )-spaces, introduced by Grothendieck [188]. Clearly, any (LB)-space is a (DF )-space. One can ask, among other things, for which (LF )-spaces E their Mackey dual (E,μ(E,E)) or strong dual (E,β(E,E)) is K-analytic or even analytic. It was known already [83] that the precompact dual of any separable (LF )- space is analytic. In [367], [368], the class of strongly weakly compactly generated (SWCG) Banach spaces E for which the Mackey topology μ(E,E) arises in a natural way, was introduced. There are many interesting topological problems related to the classes of lcs’s above. Let us mention a few of them strictly connected with the topic of the book. Floret [166], motivated by earlier works of Grothendieck, Fremlin, De Wilde and Pryce, asked if the compact sets in any (LF )-space are metrizable and if any (LF )- space is weakly angelic. Although the first question (as we have already mentioned in the Preface) has been answered positively for (DF )-spaces and dual metric spaces, both questions for (LF )-spaces were solved (also positively) by Cascales and Orihuela in [81]. Orihuela [320] answered the second question also for dual metric spaces. Therefore, it was natural to ask about a possible large class of lcs (clearly including (LF )-spaces and dual metric spaces) for which both questions also have positive answers. Such a class of lcs, called the class G, was introduced by Cascales and Orihuela [82]. An lcs E belongs to G if the weak∗ dual E admits a N resolution {Kα : α ∈ N } consisting of σ(E ,E) relatively countably compact sets such that each sequence in any Kα is equicontinuous. Spaces in the class G enjoy interesting topological properties, such as: 10 1Overview

(i) The weak topology of E ∈ G is angelic, and every precompact set in E is metrizable. (ii) For E ∈ G, the densities of E and (E,σ(E,E)) coincide if (E, σ (E, E)) is a Lindelöf Σ-space [82]; this extends a classical result of Amir and Lindenstrauss for WCG Banach spaces. (iii) For a compact space X, the space Cp(X) is K-analytic if and only if it is homeomorphic to a weakly compact set of a locally convex space in the class G, see [82]. Recall that a compact space X is Eberlein compact if and only if it is homeomorphic to a weakly compact subset of a Banach space. (iv) An lcs in G is metrizable if and only if it is Fréchet–Urysohn. A barrelled lcs E in the class G (for example, any (LF )-space E) is metrizable if and only if E is Fréchet–Urysohn, if and only if E is Baire-like if and only if E does not contain φ (i.e., the ℵ0-dimensional vector space with the finest locally convex topology [89]). (v) Every quasibarrelled space E in G has countable tightness, and the same also holds true for (E, σ (E, E)) [88]. This extends a classical results of Kaplansky; see [165]. On the other hand, there is a large and important class of lcs that does not belong to G.AnlcsCp(X) belongs to the class G if and only if X is countable (i.e., Cp(X) is metrizable); see [89]. Therefore, many results for spaces Cp(X) (also presented in the book) require methods and techniques different from those applied to study the class G; we refer the reader to the excellent works about Cp(X) theory in [25] and [24]. Recall that a topological space X is sequential if every sequentially closed set in X is closed. Clearly, from the definition, we have metrizable ⇒ Fréchet–Urysohn ⇒ sequential ⇒ k-space. Probably the first proof that an (LF )-space is metrizable if and only if it is Fréchet–Urysohn was presented in [224]. Cascales and Orihuela’s [81] result that (LF )-spaces are angelic proved that any (LF )-space is sequential if and only if it is a k-space. Nyikos [316] observed that the (LB)-space φ is sequential and not Fréchet– Urysohn. Much earlier, Yoshinaga [436] had proved that the strong dual of any Fréchet–Schwartz space (equivalently, every Silva space) is sequential. Next, Webb [415] extended this result to strong duals of Fréchet–Montel spaces (equivalently, to Montel-(DF )-spaces). He proved that these spaces are Fréchet–Urysohn only when finite-dimensional. Since Montel (DF )-spaces form a part of the class of (LB)-spaces, one can ask if the Nyikos–Yoshinaga–Webb result extends further within (LB)-spaces or (DF )- spaces. It turns out [229] that the answers are negative. The strong dual of a metrizable lcs E is sequential if and only if E is a dense subspace of either a Banach space or a FréchetÐMontel space. Apparently, any proper (LB)-space is not Fréchet–Urysohn since proper (LB)- spaces contain a copy of φ;see[356]. As we have already mentioned, Fréchet– Urysohn (DF )-spaces or (LF )-spaces are metrizable since they belong to the class G. At this point, there emerges a disparity. Namely, Webb [415] introduced the property C3 (sequential closure of each set is sequentially closed), which characterizes metrizability for (LF )-spaces; [224] but not for (DF )-spaces; see [229,As- − sertions 5.2 and 5.3]. A variant property C3 (the sequential closure of every linear 1.1 General comments and historical facts 11 subspace is sequentially closed), defined in [229], characterized both the barrelled (DF )-spaces and (LF )-spaces as being of the form M, φ,orM × φ, where M is a metrizable lcs. The main result of [229], characterizing both (DF )-spaces and (LF )-spaces that are sequential as being either metrizable or Montel (DF )-spaces, provides an answer to topological group questions of Nyikos [316, Problem 1]. Chapter 2 Elementary Facts about Baire and Baire-Type Spaces

Abstract This chapter contains classical results about Baire-type conditions (Baire-like, b-Baire-like, CS-barrelled, s-barrelled) on tvs. We include applications to closed graph theorems and C(X) spaces. We also provide the ﬁrst proof in book form of a remarkable result of Saxon (extending earlier results of Arias de Reyna and Valdivia), that states that, under Martin’s axiom, every lcs containing a dense hyperplane contains a dense non-Baire hyperplane. This part also contains analytic characterizations of certain completely regular Hausdorff spaces X. For example, we show that X is pseudocompact, is Warner bounded, or Cc(X) is a (df )-space if and only if for each sequence (μn)n in the dual Cc(X) there exists a sequence (tn)n ⊂ (0, 1] such that (tnμn)n is weakly bounded, strongly bounded, or equicontinuous, respectively. These characterizations help us produce a (df )-space Cc(X) that is not a (DF )-space, solving a basic and long-standing open question.

2.1 Baire spaces and Polish spaces

Let A be a subset of a nonvoid Hausdorff topological space X. We shall say that A is nowhere dense (or rare) if its closure A has a void interior. Clearly, every subset of a nowhere dense set is nowhere dense. A is called of first category if it is a countable union of nowhere dense subsets of X. Clearly, every subset of a first category set is again of first category. A is said to be of second category in X if it is not of first category. If A is of second category and A ⊂ B, then B is of second category. The classical Baire category theorem states the following.

Theorem 2.1 If E is either a complete metric or a locally compact Hausdorff space, then the intersection of countably many dense, open subsets of X is dense in E.

Proof We show only that the intersection of countably many dense open sets in every metric complete space (E, d) is nonvoid. If this were false, then we would = have E n En, where each En is a closed subset with empty interior. Hence there exists x1 and 0 <ε1 < 1 such that B(x1,ε1) ⊂ E \ E1, where B(x1,ε1) is the open −1 −1 ball at x1 with radius ε1. Next there exists x2 ∈ B(x1, 2 ε1) and 0 <ε2 < 2 ε1

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 13 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_2, © Springer Science+Business Media, LLC 2011 14 2 Elementary Facts about Baire and Baire-Type Spaces such that B(x2,ε2) ⊂ E \E2. Continuing this way, one obtains a shrinking sequence −n of open balls B(xn,εn) with radius less than 2 disjoint with En. Clearly, (xn)n is a Cauchy sequence in (E, d), so it converges to x ∈ E \ En, n ∈ N, and we have reached a contradiction. Similarly, one gets that each open subset of (E, d) is of second category.

This deep theorem is a principal one in analysis and topology, providing many applications for the closed graph theorems and the uniform boundedness theorem. A topological space X is called a Baire space if every nonvoid open subset of X is of second category (equivalently, if the conclusion of the Baire theorem holds). Clearly, every Baire space is of second category. Although there exist topological spaces of second category that are not Baire spaces, we note that all tvs’s considered in the sequel are assumed to be real or complex if not speciﬁed otherwise.

Proposition 2.1 If a tvs E is of second category, E is a Baire space.

Proof Let A be a nonvoid open subset of E.Ifx ∈ A, then there exists a balanced + ⊂ = neighborhood of zero U in E such that x U A. Since E n nU and E is of second category, there exists m ∈ N such that mU is of second category. Then U is of second category, too. This implies that x + U is of second category and A, containing x + U, is also of second category.

We shall also need the following classical fact; see [328, 10.1.26]. For a completely regular Hausdorff space X,byCc(X) and Cp(X) we denote the space of real-valued continuous functions on X endowed with the compact-open and pointwise topologies, respectively.

Proposition 2.2 Let X be a paracompact and locally compact topological space. Then Cc(X) is a Baire space.

Proof Since X is a paracompact locally compact space, X can be represented as the topological direct sum of a disjoint family {Xt : t ∈ T } of locally compact σ - compact spaces Xt , and we have a topological isomorphism of Cc(X) and the product t∈T Cc(Xt ). It is known that each space Cc(Xt ) is a Fréchet space (i.e., a metrizable and complete lcs) and since products of Fréchet spaces are Baire spaces (see Theorem 14.2 below for an alternative proof), we conclude that t∈T Cc(Xt ) is a Baire space.

The following general fact is a simple consequence of deﬁnitions above.

Proposition 2.3 If E is a tvs and F is a vector subspace of E, then F is either dense or nowhere dense in E. If F is dense in E and F is Baire, then E is Baire.

Proof Assume that F is not dense in E.LetG be its closure in E, a proper closed subspace of E.IfG is not nowhere dense in E, then there exists a bal- 2.1 Baire spaces and Polish spaces 15 anced neighborhood of zero U in E and a point x ∈ G such that x + U ⊂ G. Then = ⊂ E n nU G, providing a contradiction. The other part is clear.

Recall that every Cech-completeˇ space E (i.e., E can be represented as a countable intersection of open subsets of a compact space) is a Baire space. Arkhangel’skii proved [31] that if E is a topological group and F is a Cech-completeˇ subspace of E, then either F is nowhere dense in E or E is Cech-completeˇ as well. This, combined with Proposition 2.3, shows that if a tvs E contains a dense Cech-ˇ complete vector subspace, then E is Cech-complete.ˇ A subset A of a topological space X is said to have the Baire property in X if there exists an open subset U of X such that U \ A and A \ U are of ﬁrst category. Let D(A) be the set of all x ∈ X such that each neighborhood U(x) of x intersects A in a set of second category. Set O(A) := int D(A).

Proposition 2.4 A subset A of a topological space X has the Baire property if and only if O(A)\ A is of ﬁrst category.

Proof Assume A has the Baire property, and let U be an open set in X such that U \ A and A \ U are sets of ﬁrst category. Note that D(A) ⊂ U. Indeed, if x ∈ D(A) \ U, then (by deﬁnition) the set

(X \ U)∩ A(= A \ U) is of second category. By the assumption, A \ U ⊂ A \ U is of ﬁrst category, a contradiction. Since U \ U is nowhere dense, one concludes that U \ A is of ﬁrst category. Finally, since

O(A)\ A ⊂ D(A) \ A ⊂ U \ A, then O(A)\ A is of first category as claimed. Now assume that O(A) \ A is a set of first category. It is enough to prove that A \ O(A) is of first category. Let C(A) be the union of the family L := {Ai : i ∈ I} of all the open subsets of X that intersect A in a set of first category. Note that O(A) = X \ C(A). We show that

A ∩ C(A) (= A \ O(A))

{ : ∈ } is of first category. Let Ai i J be a maximal pairwise disjoint subfamily of L. ∩ ∩ Then (as is easily seen) A ( i∈J Ai) is of first category. Then the set A i∈J Ai ⊂ is also of first category. By the maximality condition, we deduce that i∈I Ai i∈J Ai , which completes the proof.

Since A ∩ C(A) is of the ﬁrst category, we note the following simple fact; see, for example, [421,p.4]. 16 2 Elementary Facts about Baire and Baire-Type Spaces

Proposition 2.5 Let E be a topological space, and let B be a subset of E that is \ { : ∈ N} the union of a sequence (Un)n of subsets of E. Then D(B) O(Un) n is nowhere dense. Therefore O(B)\ {O(Un) : n ∈ N} is also nowhere dense. Proof Assume that the interior A of the closed set D(B) \ {O(Un) : n ∈ N} is non-void. Then A ∩ B is of second category. Hence there exists m ∈ N such that A ∩ Um is of second category. Therefore, as Um ∩ C(Um) is of ﬁrst category, we have A ⊂ C(Um), and hence A ∩ O(Um) is nonvoid, a contradiction.

Every Borel set in a topological space E has the Baire property. This easily follows from the following well-known fact; see [371].

Proposition 2.6 Let E be a topological space. The family of all subsets of E with the Baire property forms a σ -algebra.

Now we are ready to formulate the following useful fact.

Proposition 2.7 Let U be a subset of a topological vector space E. If U is of second category and has the Baire property, U − U is a neighborhood of zero.

Proof Since U is of second category, we have O(U) = ∅.Ifx/∈ U −U, then clearly (x + U)∩ U =∅. The Baire property of U implies that (x + O(U)) ∩ O(U) is a set of ﬁrst category. On the other hand, since nonvoid open subsets of O(U) are of second category, it follows that (x + O(U))∩ O(U) =∅. Then x/∈ O(U)− O(U). This proves that U − U contains the neighborhood of zero O(U)− O(U).

It is clear that the fact above has a corresponding variant for topological groups (called the Philips lemma); see, for example, [346]. In general, even the self-product X ×X of a Baire space X need not be Baire; see [324], [106], [164]. Nevertheless, the product i∈I Xi of metric complete spaces is { : ∈ } Baire; see [328]. Also, the product i∈I Xi of any family Xi i I of separable Baire spaces is a Baire space; see [421]. Arias de Reina [16] proved the following remarkable theorem.

2 Theorem 2.2 (Arias de Reina) The Hilbert space (ω1) contains a family {Xt : t<ω1} of different Baire subspaces such that for all t,u < ω1, t = u, the product Xt × Xu is not Baire.

Valdivia [424] generalized this result by proving the same conclusion in each p space c0(I) and (I), for uncountable set I , and 0

Proposition 2.8 (i) The intersection of any countable family of Polish subspaces of a topological space E is a Polish space. (ii) Every open (closed) subspace V of a Polish space E is a Polish space. Hence a subspace of a Polish space that is a Gδ-set is a Polish space. := Proof (i) Let (En)n be a sequence of Polish subspaces of E, and let G n En. Then the product n En (endowed with the product topology) and the diagonal ⊂ Δ n En (as a closed subset) are Polish spaces. Since Δ is homeomorphic to the intersection G, the conclusion follows. (ii) Let d be a complete metric on E.LetV be open and V c := E \ V . Deﬁne the function d(x,V c) by d(x,V c) := inf {d(x,y) : y ∈ V c}. Set ξ(x):= d(x,V c)−1 and D(x,y) := |ξ(x)− ξ(y)|+d(x,y) for all x,y ∈ V . It is easy to see that D(x,y) deﬁnes a complete metric on V giving the original topology of E restricted to V . Hence V is a Polish space.

The following characterizes Polish subspaces of a Polish space; see [371].

Proposition 2.9 A subspace F of a Polish space E is Polish if and only if F is a Gδ-set in E.

Proof If F is a Gδ-set in E, then F is a Polish space by the previous proposition. To prove the converse, let d (resp. d1) be a compatible (resp. complete) metric on −1 E (resp. F ). For each x ∈ F and each n ∈ N, there exists 0 km, one gets ≤ + −1 + −1 + d(xn,xm) d(xn,y) d(y,xm)

−1 By construction, the inequality d(xn,xm)

Clearly, (xn)n is Cauchy in (F, d1), and from the completeness it follows that y ∈ F .

Corollary 2.1 A topological space E is a Polish space if and only if it is homeo- N morphic to a Gδ-set contained in the compact space [0, 1] .

Proof Since the space [0, 1]N is a metric compact space, it is a Polish space and Proposition 2.8 applies. To get the converse, assume that E is a Polish space. Hence it is separable and metrizable, and consequently E is homeomorphic to a subspace of [0, 1]N. Proposition 2.9 applies to complete the proof. 18 2 Elementary Facts about Baire and Baire-Type Spaces

2.2 A characterization of Baire topological vector spaces

The following characterization of a Baire tvs is due to Saxon [353]; see also [328, Theorem 1.2.2].

Theorem 2.3 (Saxon) The following are equivalent for a tvs E. (i) E is a Baire space. (ii) Every absorbing balanced and closed subset of E is a neighborhood of some point.

The scalar ﬁeld K is either the reals or the complexes. After a little preparation motivated by annular regions in the complex plane, we present a single proof that simultaneously solves both the real and complex cases. − = ∞ −1 Claim 2.1 Let U be a nonempty set in a tvs E.IfU U E, then n=1 n U has an empty interior. ∈ \ − −1 Proof Let x E (U U).IfV is a nonempty open set contained in n n U, then the open neighborhood of zero V − V contains m−1x for some sufﬁciently large m ∈ N and is contained in n−1 (U − U) for every n ∈ N. In particular, m−1x ∈ m−1 (U − U), which implies x ∈ U − U, a contradiction.

K For 0

For each ε>0, let βε denote the open ball {t ∈ K : |t| <ε}.For0<ε<1, compactness provides a ﬁnite subset Γε of Λε,1 such that + ⊃ Γε βε2 Λε,1.

Claim 2.2 If 0 <δ<ε<1 and z + βε ⊂ Λε,1, then Γδ · (z + βε) ⊃ Λδ,ε.

Proof Γδ · (z + βε) = t · (z + βε) = t · z + βε|t| ⊃ (t · z + βεδ) t∈Γδ t∈Γδ t∈Γδ = + · ⊃ + · ⊃ · = ⊃ Γδ βδε/|z| z Γδ βδ2 z Λδ,1 z Λδ|z|,|z| Λδ,ε.

Claim 2.3 Let (Bn)nbe a sequence of subsets of a tvs E.Fix00 such that ty 0 < t δ −1 n n Bn.

· ⊂ −1 Proof For each n,wehaveΛr1/n,r2/n y n Bn. There is a natural number m −1 such that r1/n ≤ r2/ (n + 1) for every n ≥ m. The claim follows for δ = r2m . 2.2 A characterization of Baire topological vector spaces 19

Now we are ready to prove Theorem 2.3.

Proof Only the implication (ii) ⇒ (i) needs a proof. Indeed, if there exists an ab- = sorbing, balanced and nowhere dense set B, then E n nB is of first category and not Baire. We assume that E is not Baire and construct such a set B. Since E is of first category, its topology is nontrivial and it contains a closed balanced neighborhood U of zero with U − U = E. Since U is also of first category in E, there exists a sequence (An)n of closed nowhere dense sets in E whose union is U. With the notation of Claim 2.2, we observe that each set Bn := Γ1/k · Aj j,k≤n is closed and nowhere dense, being a finite union of such sets. Furthermore, each Bn is contained in the balanced set U. := −1 We wish to see that A n n Bn is nowhere dense. Suppose some nonempty −1 open set W is contained inA. Because each finite union n0 such that the set βr · y is contained in U and is thus covered by (An)n. We may harmlessly assume that r ≤ 1. The set βr · y either has the trivial topology or is a topological copy of the subset βr of K. Either way, βr ·y is of second category in itself, and there exist p ∈ N, ε>0 and z ∈ βr such that

z + βε ⊂ βr and (z + βε) · y ⊂ Ap.

In fact, rechoosing z and ε if needed, we may additionally insist that z = 0, and then we may yet again reﬁne the choice of ε in the interval (0, 1) so that

z + βε ⊂ Λε,1 and (z + βε) · y ⊂ Ap.

Let q be a natural number larger than both p and ε−1.Forn ≥ q, we apply Claim 2.2 to obtain

Bn ⊃ Γ1/q · Ap ⊃ Γ1/q · (z + βε) · y ⊃ Λ1/q,ε · y. Now Claim 2.3 shows that, for some δ>0, −1 {ty : 0 < |t| ≤ δ} · y ⊂ n Bn ⊂ A. n≥q

Therefore A absorbs y, given that zero is (obviously) in A. The balanced core of A (i.e., the largest balanced set A0 contained in A) is absorbing and nowhere dense because A has these properties. Therefore B := A0 is a closed, balanced, absorbing nowhere dense set in E, as promised. 20 2 Elementary Facts about Baire and Baire-Type Spaces

Theorem 2.3 provides the following corollary.

Corollary 2.2 Every Hausdorff quotient of a Baire tvs is a Baire space.

We also have the following useful corollary.

Corollary 2.3 If a Baire tvs E is covered by a sequence (En)n of vector subspaces of E, then Em is dense and Baire for some m ∈ N.

Proof By hypothesis there exists m ∈ N such that Em is of second category in E; therefore it cannot be nowhere dense in E. By Proposition 2.3, Em is dense in E, thus of second category in itself and thus Baire by Proposition 2.1.

When dim(E) is inﬁnite, Em may satisfy dim(E/Em) = dim(E),anextreme.

Proposition 2.10 Every inﬁnite-dimensional Baire tvs E contains a dense Baire subspace F whose dimension equals the codimension in E.

Proof By (xt )t∈T denote a Hamel basis of E. Fix a partition (Tn)n of T such that = ∈ N := { : ∈ n } card T card Tn for all n . Set En span xt t i=1 Ti . Then (En)n covers E and dim E = dim En = dim(E/En) for n ∈ N. By Corollary 2.3, there exists a dense Baire subspace F := Em of E, as desired.

At the other algebraic extreme, hyperplanes of E are also Baire when closed, and those that contain F are dense and Baire.

2.3 Arias de Reyna–Valdivia–Saxon theorem

For many years, the following question remained: When does an inﬁnite-dimensional Baire tvs E admit a non-Baire (necessarily dense) hyperplane? In 1966, Wilansky and Klee conjectured: Never, for E a Banach space. This conjecture was denied in 1980 by Arias de Reyna [328, Theorem 1.2.12], who proved, under Martin’s axiom, the answer: Always, when E is a separable Banach space. In 1983, Valdivia [427] proved, under Martin’s axiom, the more general answer: Always, when E is a separable tvs. In 1987, Pérez Carreras and Bonet [328, Question 13.1.1] repeated the question for E a (not necessarily separable) Banach space. Finally, in 1991, Saxon [355] provided a complete answer in the general locally convex setting. He proved the following theorem.

Theorem 2.4 (Arias de ReynaÐValdiviaÐSaxon) Assume c-A. Every tvs E with an inﬁnite-dimensional dual contains a non-Baire hyperplane. Consequently,(1)every inﬁnite-dimensional lcs admits a non-Baire hyperplane and (2) an lcs E admits a dense non-Baire hyperplane iff E = E∗. 2.3 Arias de ReynaÐValdiviaÐSaxon theorem 21

For a tvs E, by the dual of E we mean its topological dual E, a linear subspace of its algebraic dual E∗. A subset A of a tvs E is called bornivorous if A absorbs every bounded set in E. Recall that an lcs E is barrelled (quasibarrelled) if every closed absolutely convex and absorbing (and bornivorous) subset of E is a neighborhood of zero of E or, equivalently, if every bounded set in the weak dual (E,σ(E,E)) (strong dual (E,β(E,E))) is equicontinuous. ℵ The axiom c-A (c-additivity, where c := 2 0 ) proclaims: The union of less than c subsets of R, each of measure zero, itself has measure zero. Note that CH ⇒ Martin’s axiom ⇒ c-A, and the converse implications fail in general; see [172, Corollary 32(G)(c)]. To prove Theorem 2.4, we will need the following technical fact from [328, The- orem 1.2.11].

Lemma 2.1 Let e1 and e2 be the canonical unit vectors in the real Euclidean space R2 endowed with its usual inner product (., .) and the corresponding norm .. By −1 m(x) := arccos ((x, e1)x ), for x = 0, we denote the angle between x and e1. If 0 q−1 and u − v

Now we prove Theorem 2.4.

Proof It is enough to prove the initial claim. Clearly, (1) then follows and (2) as well. Indeed, if E is an lcs with E = E∗, then E contains a non-Baire hyperplane H by (1). If (a) E is Baire, then all its closed hyperplanes are Baire, and H must be dense. If (b) E is non-Baire, then so are all its hyperplanes, including dense hyperplanes, which exist by the hypothesis E = E∗. Conversely, no dense hyperplanes exist if E = E∗. We prove the real scalar case only. One may then easily dispatch the complex case by a standard procedure. We also assume, without loss of generality, that E is Baire. The hypothesis implies a biorthogonal sequence (xn,hn)n ⊂ E × E ,sothemap N T : E −→ R deﬁned by T (x) := (hn (x))n is continuous and linear. Therefore E admits a quotient F := E/Q of dimension at most c isomorphic to a subspace of RN containing the canonical unit vectors and endowed with a vector topology ﬁner than the one inherited from the usual product topology. Hence each unit vector en and each coordinate functional fn belong to F and F , respectively. For each n∈ N,setMn := span{ei : i ≤ n}.If(wn,k)k is an enumeration of the countable set { n : ∈ Q = } ∈ N i=1 aiei ai ,an 0 , then (wn,k)k is dense in Mn for all n . Set i −i Ui := {x ∈ F :|fn(x)| < 2 } n=1 for each i ∈ N. Then Ui ={0},Ui+1 + Ui+1 ⊂ Ui,i∈ N. (2.1) i 22 2 Elementary Facts about Baire and Baire-Type Spaces

For each ﬁxed n ∈ N, we choose a sequence (εn,k)k of numbers such that

−n 0 < 2εn,k+1 ≤ εn,k < 2 ∧|fn(wn,k)|.

Note that i wi,k ∈/ Vi,k := {x ∈ F :|fn(x)| <εi,k}⊂Ui n=1 for all i, k ∈ N. Moreover,

Vn,k+1 + Vn,k+1 ⊂ Vn,k ⊂ Un,n,k∈ N (2.2) and Ln := (wn,k + Vn,k) ⊂ F \{0},n∈ N. (2.3) k = ⊕ ⊥ Each Ln is dense and open in F( Mn ( i≤n fi )). To complete the proof, we need only find a non-Baire hyperplane G in F ; indeed, the hyperplane H in E satisfying H ⊃ Q and G = H/Q would also be non-Baire by Corollary 2.2.Note that ℵ0 ≤ dim (F ) =: α ≤ c. From the previous section, we know that the Baire space F contains a dense Baire hyperplane P .Let B := yβ : β is an ordinal <α be a Hamel basis for P .IfA is the absolutely convex envelope of B, then A ∩ P is a barrel in the dense barrelled subspace P and, consequently, A is a neighborhood of zero in F . Hence we can find a point z ∈ F \ P such that 2z ∈ A. We observe from \ := (2.3) that the set F L is of first category in F , where L n Ln. The proof will be complete if we find a dense hyperplane G contained in F \ L. The space G will take the form

G := span{yγ + aγ z : γ<α} for scalars aγ suitably chosen with each |aγ |≤1. Formula (2.3) ensures that L misses

{0}=span ∅=span{yγ + 1 · z : γ is an ordinal < 0}.

Zorn’s lemma provides a maximal subspace M of F of the form

M := span{yγ + aγ z : γ<β} subject to the conditions that L misses M, the ordinal β does not exceed α and each |aγ |≤1.

Claim 2.4 We have β = α. 2.3 Arias de ReynaÐValdiviaÐSaxon theorem 23

Indeed, assume that α = β.Theset{γ : γ<β} and the family F of all its ﬁnite subsets have less than c elements. Now we will apply Lemma 2.1 to our situation. 2 Identify yβ and z with the unit vectors e1 and e2, respectively, and R with the linear span X of yβ and z. Let h : M + X → X be the projection onto X along M.LetD be the family of all subspaces of the form X + span{yγ + aγ z : γ ∈ J } for J ∈ F. Note that D covers M + X and |D| < c. Next observe that L ∩ (M + X) = LD,n, n∈N,D∈D

−1 where LD,n := L∩D ∩{x :h(x) >n }. This follows from the fact that h(x) = 0 if x ∈ L ∩ (M + X), because then x/∈ M (since M misses L). Fix arbitrary n ∈ N and D ∈ D as well as 0 1 such that −1 h(x)

−1 LD,n ⊂ Lr ∩ D ∩{x :h(x) >n }= −1 D ∩ (wr,k + Vr,k ) ∩{x :h(x) >n }. k Assume for the moment that x and y are in the kth set. Then h(x) >n−1 and −1 k−1 h(y) >n . Moreover, 2 (x − y) ∈ D. Since x − y ∈ Vr,k + Vr,k ,weuse(2.2) and (2.3) and k − 1 more steps to obtain

k−1 2 (x − y) ∈ Vr,1 + Vr,1 ⊂ Ur + Ur ⊂ Ur−1.

Hence, by continuity of h|D,wehave − − h(x − y) < 21 ktn 1.

Now we can apply Lemma 2.1 to obtain that − |m(h(x)) − m(h(y))| < 22 kt.

This proves that the set m(h(LD,n))is covered by a sequence of intervals whose = 2−k union has measure less than 4t k 2 t. Since the scalar t was arbitrary, we conclude that m(h(LD,n)) has measure zero. Now axiom c-A declares that

C := m(h(L ∩ (M + X))) has measure zero. Therefore we can ﬁnd an angle θ such that 0 <θ<π/4 and θ/∈ C and π − θ/∈ C.Nowsetaβ := tan θ and v = yβ + aβ z. Hence 0

m(h(x + bv)) = m(bv) = m(v) = θ, m(h(x − bv)) = m(−v) = π − θ 24 2 Elementary Facts about Baire and Baire-Type Spaces if x ∈ M and b>0. This implies that x ± bv∈ / L. We know already that M misses L.NowM + span{v} also misses L, which contradicts maximality of M, proving the claim. Finally, we prove that G := M is dense in F . Since G is a hyperplane, it is enough to show that G is not closed. Now tz∈ / M for each t = 0. Since 2z ∈ A, there exists a net (z ) = ( b y ) in A that converges to 2z, where J is finite and u u γ ∈Ju u,γ γ u u |b |≤1. The corresponding net ( b a z) has an adherent point γ ∈Ju u,γ γ ∈Ju u,γ γ u pz in the compact interval {bz :|b|≤1}.LetU and V be neighborhoods of 2z and pz, respectively. V contains a cofinal subnet of the second net, and U contains points of the corresponding cofinal subnet of the first net. Hence U + V contains points of a subnet of ( b (y + a z)) ⊂ M. It follows that 2z + pz ∈ M\M, and γ ∈Ju u,γ γ γ u G = M is not closed. The proof is complete.

Remark 2.1 If E is a tvs with a separable quotient F such that F = F ∗, one can ﬁnd points wn,k in F and balanced open neighborhoods Ui and Vi,k of zero satisfying (2.1) and (2.2) such that each Ln deﬁned as in (2.3) is dense (and open) in F . It then follows from the proof of Theorem 2.4 that, under the assumption of c-A, there exists in E a dense non-Baire hyperplane.

2.4 Locally convex spaces with some Baire-type conditions

A new line of research concerning Baire-type conditions started with the AmemiyaÐ Komuraø theorem (see [328, Theorem 8.2.12] and [419]), stating that if (An)n is an increasing sequence of absolutely convex closed subsets covering a metrizable and barrelled lcs E, then there exists m ∈ N such that Am is a neighborhood of zero in E. Saxon [354], motivated by the AmemiyaÐKomuraø result, deﬁned an lcs E to be Baire-like if for any increasing sequence (An)n of closed absolutely convex subsets of E covering E there is an integer n ∈ N such that An is a neighborhood of zero. If the sequence (An)n is required to be bornivorous (i.e., for every bounded set B in E there exists Am that absorbs B), then Ruess deﬁnes E to be b-Baire-like. Clearly, for an lcs, Baire implies Baire-like, Baire-like implies b-Baire-like and barrelled, and (b)-Baire-like implies (quasi)barrelled. The main purpose of the research started by Saxon was to study stable lcs properties inherited by products and small-codimensional subspaces of Baire spaces. The Baire-like property is such an example. Although products [424] and countable- codimensional subspaces (even hyperplanes) [16] of Baire locally convex spaces need not be fully Baire, they are always Baire-like since countable-codimensional subspaces of Baire-like spaces are Baire-like and topological products of Baire-like spaces are Baire-like; see [354]. These weak Baireness/strong barrelledness properties have well occupied other authors and books; see [421], [328], [260], [162], [356], [362], [361]. Notwithstand- ing, our selection and treatment is unique. 2.4 Locally convex spaces with some Baire-type conditions 25

Metrizable barrelled spaces are Baire-like (AmemiyaÐKomura)ø and include all metrizable (LF )-spaces [328, Proposition 4.2.6]. Conversely, as we shall soon see, Baire-like (LF )-spaces must be metrizable. (i) Let E be a vector space, and let (En,τn)n be an increasing sequence of vector subspaces of E covering E, each En endowed with a locally convex topology τn, such that τn+1|En ≤ τn for each n ∈ N. Then on E there exists the finest locally convex topology τ such that τ|En ≤ τn for each n ∈ N.Ifτ is Hausdorff, we say that (E, τ) is the inductive limit space of the sequence (En,τn)n, and the latter is a defining sequence for (E, τ). (ii) If each (En,τn) is metrizable, then (E, τ) is an (LM)-space. (iii) If each (En,τn) is a Fréchet space (a Banach space), then (E, τ) is an (LF )- space (an (LB)-space). (iv) If τn+1|En = τn for each n ∈ N, then τ|En = τn for each n ∈ N, (E, τ) is the strict inductive limit of (En,τn)n and (E, τ) is complete if each (En,τn) is complete; see [328, Proposition 8.4.16]. (v) An (LF )-space E is called proper if it has a defining sequence of proper subspaces of E. The Bairelikeness of some concrete normed vector-valued function spaces was studied by several specialists; see [129], [130] for details. We note only that the normed spaces of Pettis or Bochner integrable functions are not Baire spaces but Baire-like; see also [420] for more examples of normed Baire-like spaces that are not Baire. We also have the following simple proposition.

Proposition 2.11 Every metrizable lcs E is b-Baire-like.

Proof Let (Un)n be a decreasing base of absolutely convex neigborhoods of zero in E. Assume that E is not b-Baire-like. Then there exists a bornivorous sequence (An)n of absolutely convex closed sets such that Un ⊂ nAn for each n ∈ N. Choose

xn ∈ Un \ nAn.

Since the null sequence (xn)n is bounded, it is contained in mAk for some k,m ∈ N. Hence, for all n ≥ max{m, k},wehave

{xn : n ∈ N}⊂mAk ⊂ nAn, a contradiction.

Proposition 2.12 Every barrelled b-Baire-like space E is Baire-like.

Proof Let (An)n be an increasing sequence of absolutely convex closed subsets of E covering E. The proof will be ﬁnished if we show that (An)n is bornivorous. Indeed, then we apply that E is b-Baire-like, and some Am will be a neighborhood of zero. Assume, by way of contradiction, that there exists a bounded set B ⊂ E such that B ⊂ nAn for each n ∈ N. For each n ∈ N, select −1 xn ∈ n B \ An. 26 2 Elementary Facts about Baire and Baire-Type Spaces

Since each set An is closed, for each n ∈ N there exists a closed and absolutely convex neighborhood of zero Un such that

Un+1 + Un+1 ⊂ Un,xn ∈/ An + Un.

Set U := An + Un. n Then U is a barrel in E and is a neighborhood of zero to which almost all elements of the null sequence (xn)n must belong, a contradiction.

Proposition 2.12 shows that every metrizable barrelled space is Baire-like, the AmemiyaÐKomuraø result. A more general fact, due to Saxon, is known [354, The- orem 2.1]: A barrelled lcs that does not contain a (isomorphic) copy of ϕ (i.e., an ℵ0-dimensional vector space with the ﬁnest locally convex topology) is Baire-like. Saxon actually proved the following [354, Corollary 2.2], which is even a bit more general; see also [222].

Theorem 2.5 Let E be ∞-barrelled (i.e., every σ(E,E)-bounded sequence in E is equicontinuous). Assume that E is covered by an increasing sequence (An)n of absolutely convex closed subsets of E such that no set An is absorbing in E. Then E contains a copy of ϕ.

Proof By considering subsequences, we may assume that the span of each An is a proper subspace of the span of An+1. Arbitrarily select (xn)n in E such that xn ∈ An+1 \ span(An). We show that the span S of the necessarily linearly independent sequence (xn)n is a copy of ϕ. It sufﬁces to show that if p is an arbitrary seminorm on S, there exists a continuous seminorm q on E such that p ≤ q|S. Without loss of generality, we may assume that ⎛ ⎞ n n ⎝ ⎠ p aj xj = |aj |p(xj ), p(xn) ≥ 1,n∈ N. j=1 j=1

We proceed inductively to ﬁnd a sequence (fn)n in E such that ∈ ◦ fn An, (2.4) and if n ∈ N and a1,a2,...,an are scalars, then n n −n max fr aj xj ≥ (1 + 2 )p aj xj . (2.5) 1≤r≤n j=1 j=1 ∈ ◦ The HahnÐBanach separation theorem provides f1 A1 such that

−1 |f1(x1)|≥(1 + 2 )p(x1). 2.4 Locally convex spaces with some Baire-type conditions 27

Let k ∈ N, and assume there exist f1,f2,...,fk in E such that the conditions above are satisﬁed for n ≤ k. Deﬁne ⎧ ⎫ ⎨ k ⎬ −k−1 D := x = aj xj : max |fr (x + xk+1)| <(1 + 2 )p(x + xk+1) . ⎩ 1≤r≤k ⎭ j=1

If D is empty, we complete the induction step by letting fk+1 = 0. Assume that D is nonempty. For x ∈ D,wehave

−k (1 + 2 )p(x) − max |fr (xk+1)|≤ max |fr (x + xk+1)| 1≤r≤k 1≤r≤k −k−1 <(1 + 2 )[p(x) + p(xk+1)].

This yields

−k−1 −k−1 2 p(x) ≤ max |fr (xk+1)|+(1 + 2 )p(xk+1). 1≤r≤k := ∞ := + { } Hence γ supx∈D p(x) < .ForA Ak+1 span x1,...,xk ,wehave

Ak+1 ⊂ A, xk+1 ∈ span A = span Ak+1.

Moreover, A is absolutely convex and closed. The HahnÐBanach theorem provides ∈ ◦ ⊂ ◦ fk+1 A Ak+1 such that

−k−1 fk+1(xk+1) = (1 + 2 )(γ + p(xk+1)).

Thus x ∈ D implies that

−k−1 −k−1 fk+1(x + xk+1) = (1 + 2 )(γ + p(xk+1)) ≥ (1 + 2 )p(x + xk+1). = + = + ∈ S To prove (2.5)forn k 1, consider an arbitrary element z y axk+1 = k with y j=1 aj xj and

−k−1 |fr (z)| <(1 + 2 )p(z) for 1 ≤ r ≤ k. By the induction assumption, a = 0 and a−1y ∈ D. Therefore, by the above, we have −1 −k−1 −1 |fk+1(a z)|≥(1 + 2 )p(a z). Thus (2.4) and (2.5) hold for n ≤ k + 1; the induction is complete. Now ﬁx x ∈ E. ∈ ◦ ∈ ∈ N | |≤ Since fn An, the fact that x An for almost all n means that fn(x) 1 for almost all n ∈ N. Thus (fk)k is σ(E ,E)-bounded, and equicontinuous by hypothesis on E. The formula

x → q(x) := sup |fk(x)| k 28 2 Elementary Facts about Baire and Baire-Type Spaces deﬁnes a continuous seminorm on E.By(2.5)wehavep ≤ q|S. This shows that p is continuous on S. We proved that each seminorm on S is continuous, which shows that S is as desired.

This yields the following corollary.

Corollary 2.4 (Saxon) Every barrelled lcs that does not contain a copy of ϕ is a Baire-like space.

Hence, if a barrelled lcs E admits a ﬁner metrizable locally convex topology, then E is Baire-like. The converse implication fails. Indeed, let E = (E, ϑ) be an uncountable product of a metrizable and complete lcs. Clearly E is a nonmetrizable Baire lcs for which every ﬁner locally convex topology ξ is nonmetrizable, for if we assume that ξ has a countable base (Un)n of absolutely convex neighborhoods of ϑ zero, then, because ϑ is barrelled, (Un )n is a countable base of neighborhoods of zero for ϑ, an impossibility.

Corollary 2.5 Let E be an lcs such that each σ(E,E)-bounded sequence in E is equicontinuous and E is covered by a strictly increasing sequence of closed subspaces. Then E contains a copy of ϕ.

In fact, an ∞-barrelled space E contains a complemented copy of ϕ if and only if E is covered by a strictly increasing sequence of closed subspaces (see [356, Theorem 1]).

Corollary 2.6 Every proper (LB)-space E contains a copy of ϕ.

Proof We may assume that the unit ball Un in the nth deﬁning Banach space (En,τn) is contained in Un+1 for each n ∈ N.NoUn is absorbing in the barrelled space E. Otherwise, En would be a dense barrelled subspace of E, consequently complete by the open mapping theorem, so that En = E, a contradiction. The theorem applies with An = Un.

Hence no proper (LB)-space is metrizable. Nevertheless, metrizable and even normable proper (LF )-spaces do exist; see [328] for details and references.

Corollary 2.7 For an (LF )-space E, the following conditions are equivalent: (i) E is Baire-like. (ii) E is metrizable. (iii) E does not contain ϕ.

Proof (i) ⇒ (ii): Let (En)n be a deﬁning sequence of Fréchet spaces for E, and for each n ∈ N let Fn be a countable base of neighborhoods of zero in the Fréchet 2.4 Locally convex spaces with some Baire-type conditions 29 F F M space En.Theset of all unions of ﬁnite subsets of n n is countable. Let be the set of all absolutely convex closed neighborhoods of zero in E. Given any A ⊂ E, let acA denote the absolutely convex closed envelope of A in E.Theset

N := M ∩{acA : A ∈ F } is countable. Given an arbitrary U in M , for each n ∈ N there exists Un ∈ Fn with Un ⊂ U, and we set n An := Uj ∈ F . j=1

Clearly, the increasing sequence (n · acAn)n covers E, and each acAn ⊂ U.IfE is Baire-like, some acAn is in M and hence in N , which proves that the countable N is a base of neighborhoods of zero in E (i.e., E is metrizable). (ii) ⇒ (iii) is clear since ϕ is nonmetrizable. (iii) ⇒ (i) follows from Corollary 2.4.

We provide later a general result that in particular shows also that nonmetrizable (LF )-spaces are not Baire-like. We shall need also in the sequel the following result due to Valdivia [419]; see also [328, Proposition 8.2.27].

Proposition 2.13 (Valdivia) Let F be a dense barrelled (quasibarrelled) subspace of an lcs E, and assume that (Bn)n is an increasing sequence of absolutely convex sets such that each point (each bounded set) of F is absorbed by some Bn. Then + = = ε>0 n(1 ε)Bn n Bn, where the closure is in E. In particular, E n nBn, = and if (Bn)n is a covering of F , then E n Bn.

Proof We prove only the barrelled case. It is enough to show that Bn ⊂ (1 + ε)Bn n n ∈ + for each ε>0. If there exists ε>0 and x/ n(1 ε)Bn, we may choose for each n an absolutely convex neighborhood of zero Un in E such that

x/∈ (1 + ε) Bn + 2Un.

Set U := εBn + Un, n and observe that U ∩ F is a barrel in the barrelled space F . Thus U ∩ F is a neighborhood of zero in F , and U is a neighborhood of zero in E by density of F .But, 30 2 Elementary Facts about Baire and Baire-Type Spaces for each n,wehave

x/∈ (1 + ε) Bn + 2Un ⊃ Bn + εBn + Un ⊃ Bn + εBn + Un ⊃ Bn + U, ∈ + ∈ so that x/ n Bn U. Hence x/ n Bn.

We note the following corollary.

Corollary 2.8 For any completely regular Hausdorff space X, the space Cp(X) is b-Baire-like.

Proof Let (An)n be a bornivorous increasing sequence of absolutely convex closed subsets of Cp(X) covering Cp(X). For each n ∈ N,letBn be the closure of An in RX RX the space . Since the space Cp(X) is dense in and Cp(X) is quasibarrelled RX = = by [213, Theorem 2], we apply Proposition 2.13 to get Cp(X) n Bn. X Since R is a Baire space, and the sets Bn are closed and absolutely convex, there X exists m ∈ N such that Bm is a neighborhood of zero in R . Consequently, Am is a neighborhood of zero in Cp(X).

Corollary 2.9 If E is a barrelled space covered by an increasing sequence of absolutely convex complete subsets, then E is complete.

An lcs E is called locally complete if every bounded closed absolutely convex set B is a Banach disc (i.e., the linear span of B endowed with the Minkowski functional −1 norm xB := inf{ε>0 : ε x ∈ B} is a Banach space). An lcs E is docile if every inﬁnite-dimensional subspace of E contains an inﬁnite-dimensional bounded set; see [229]. It is known by the SaxonÐLevinÐValdivia theorem that every countable-codimensional subspace of a barrelled space is barrelled; see [328, Theorem 4.3.6]. Theo- rem 2.6 is due to Saxon; see [352]. In fact, Saxon [358] proved that even F is totally barrelled. An extension of this result to a barrelled lcs E such that (E,μ(E,E))is complete was obtained by Valdivia; see [328, Proposition 4.3.11].

Theorem 2.6 (Saxon) If F is a subspace of an (LF )-space E such that ℵ dim (E/F ) < 2 0 , then F is barrelled.

For the proof, we need the following lemma.

ℵ Lemma 2.2 (Sierpinski)« Every denumerable set S admits 2 0 denumerable subsets ℵ Sι (ι ∈ I and |I| = 2 0 ) that are almost disjoint (i.e., if ι and τ are distinct members of the indexing set I , then Sι ∩ Sτ is ﬁnite).

Proof Take S to be all rational numbers, let I be the irrationals, and for each ι ∈ I choose Sι to be a sequence of rationals that converges to ι. 2.4 Locally convex spaces with some Baire-type conditions 31

Lemma 2.3 Let A be a closed absolutely convex set having span F in a docile ℵ locally complete space E. Then dim (E/F) is either ﬁnite or at least 2 0 .

Proof Assume dim (E/F) is infinite. Then there is an infinite-dimensional subspace G of E that is transverse to F . By docility there is an infinite-dimensional bounded absolutely convex set B in G, and, trivially, its span is transverse to F .Letx1 be an arbitrary nonzero member of B. The bipolar theorem provides f1 ∈ E such that

◦ 1 f1 ∈ A and f1 (x1) = 1 · 2 . := ⊥ ∩ Now the span of B1 f1 B is inﬁnite-dimensional and transverse to the span of A1 := A + span{x1}, and A1 is absolutely convex and closed. Let x2 be an arbitrary nonzero member of B1, and use the bipolar theorem as before to obtain f2 ∈ E such that ∈ ◦ = · 2 f2 A1 and f2 (x2) 2 2 . = ⊥ ∩ = + { } By setting B2 f2 B1 and A2 A1 sp x2 and continuing inductively, we ⊂ ⊂ obtain (xn)n B and (fn)n E such that ⊂ ◦ = = = · n ∈ N (fn)n A ; fi xj 0fori j; and fn (xn) n 2 for all n . −n Since (xn)n is bounded and E is locally convex, the series n 2 xn is abso- { } lutely, and hence subseries, convergent. Let Nι ι∈I be a collection of c denumerable subsets of N that are almost disjoint (Lemma 2.2), and for every ι ∈ I set −n yι = 2 xn. n∈Nι

We claim that the set {yι}ι∈I is linearly independent and has span transverse to F . Indeed, if y is any (ﬁnite) linear combination in which the coefﬁcient of some yι is a nonzero scalar a, then we have

fn (y) = n · a for all but finitely many n ∈ Nι. ⊂ ◦ Therefore (fn)n is unbounded at y, whereas (fn)n A implies that (fn)n is bounded at all points of F since they are absorbed by A. We have thus shown { } that yι ι∈I consists of c linearly independent vectors whose span is transverse to F , as desired. This implies that every infinite-dimensional locally complete docile ℵ space E has dimension at least 2 0 . Indeed, take A = F = {0}. In particular, every ℵ infinite-dimensional Fréchet space has dimension at least 2 0 ; see also [277].

We are ready to prove Theorem 2.6.

Proof Let B be a barrel in F ,let(En)n be a defining sequence of Fréchet spaces for E,letA be the closure of B in E and let G be the span of A. In each Fréchet space ℵ En, the intersection A ∩ En is closed and has a span of codimension less than 2 0 32 2 Elementary Facts about Baire and Baire-Type Spaces in En; by Lemma 2.3, the codimension is finite. Since the countable union of finite sets is countable, the codimension of G is countable in E. Therefore G is barrelled by the SaxonÐLevinÐValdivia theorem, and A is a neighborhood of zero in G.It follows that B = A ∩ F is a neighborhood of zero in F .

The following observation will be used below; see [328, Proposition 4.1.6].

Proposition 2.14 If E is a barrelled lcs covered by an increasing sequence (En)n of vector subspaces, then a linear map f : E → F from E into an lcs F is continuous if and only if there exists m ∈ N such that the restrictions f |En : En → F are continuous for all n ≥ m, where the topology of each En is that induced by E.

There is an interesting application of Baire-like spaces for closed graph theorems; see [354]. It is known that the class of barrelled spaces is the largest class of lcs for which the closed graph (open mapping) theorem holds vis-á-vis Fréchet spaces. Bourbaki [66, Theorem III.2.1] observed that the class of barrelled spaces is also the largest one for which the uniform boundedness theorem holds. Saxon [354] showed that the Grothendieck factorization theorem for linear maps from a Baire lcs into an (LF )-space with closed graph remains true for linear maps from a Baire-like space into an (LB)-space. We have the following theorem.

Theorem 2.7 (Saxon) Let F be a Baire-like space. Let E be an (LB)-space with a deﬁning sequence (En)n of Banach spaces. Then every linear map f : F → E with a closed graph is continuous.

Proof Via equivalent norms, we may assume that the unit ball Bn for En is con- −1 tained in Bn+1 and that (Bn)n covers E. Thus (f (Bn))n covers F , and there is −1 some m ∈ N such that f (Bn) is a neighborhood of zero in F for each n ≥ m. −1 −1 Hence the closure of f (Vn) in f (En) is a neighborhood of zero whenever Vn is a neighborhood of zero in the Banach space En and n ≥ m (i.e., the restriction of −1 f to f (En) is almost continuous as a mapping into the Banach space En). Since the Banach topology is ﬁner than that induced by E, the graph of this restriction is also closed. By Ptak’s closed graph theorem [328], this mapping is continuous, and all the more so when the range space En is given the coarser topology induced by E. Since F is barrelled, Proposition 2.14 ensures that the unrestricted f is also continuous.

We complete this section with a characterization of the Bairelikeness of spaces Cc(X) for some concrete spaces X. In what follows, X is a completely regular Hausdorff topological space. If F := {f ∈ C(X) : f(X)⊂[0, 1]}, the subspace

{(f (x) : f ∈ F ) : x ∈ X} is homeomorphic to X; we identify X with this subspace, and the closure of X in [0, 1]F is the StoneÐCechˇ compactification of X, denoted by βX. Taking into 2.4 Locally convex spaces with some Baire-type conditions 33 account the restrictions to βX of the coordinate projections of [0, 1]F ,wehave f ∈ F , and therefore each uniformly bounded f ∈ C(X) has a unique continuous extension to βX. By the realcompactification υX of X, we mean the subset of βX such that x ∈ υX if and only if each f ∈ C(X) admits a continuous extension to X ∪{x}.Fromthe regularity of X, it follows that each f ∈ C(X) admits a continuous extension to υX. Therefore the closure in RC(X) of {(f (x) : f ∈ C(X)) : x ∈ X} is homeomorphic to υX. By definition, X is called realcompact if X = υX. From the continuity of the coordinate projections, it follows that X is realcompact if and only if X is homeomorphic to a closed subspace of a product of the real lines. Clearly, closed subspaces of a realcompact space are realcompact, and any product of realcompact spaces is realcompact. The intersection of a family of realcompact subspaces of a space is realcompact because this intersection is homeomorphic to the diagonal of a product. Recall that a subset A ⊂ X of a topological space X is topologically bounded if the restricted map f |A is bounded for each f ∈ C(X); otherwise we will say that A is topologically unbounded. If X is completely regular and Hausdorff, then A ⊂ X is topologically bounded if and only if for each locally finite family F the family {F ∈ F : F ∩ A = ∅} is finite. If X is a topologically bounded set of itself, then X is called pseudocompact. Recall here that an lcs E is said to be bornological if every bornivorous absolutely convex subset of E is a neighborhood of zero. The link between properties of Cc(X) and Cp(X) and topological properties of X is illustrated by Nachbin [307], Shirota [377], De Wilde and Schmets [113] and Buchwalter and Schmets [69].

Proposition 2.15 (NachbinÐShirota) Cc(X) is barrelled if and only if X is a μ- space (i.e., every topologically bounded subset of X has compact closure).

Proposition 2.16 (NachbinÐShirota, De WildeÐSchmets) The space Cc(X) is bornological if and only if Cc(X) is the inductive limit of Banach spaces if and only if X is realcompact.

Proposition 2.17 (BuchwalterÐSchmets) Cp(X) is barrelled if and only if every topologically bounded subset of X is ﬁnite.

A good sufficient condition for Cc(X) to be a Baire space is hard to locate. Let us mention the following one: If X is a locally compact and paracompact space, Cc(X) is Baire. The argument uses the well-known fact that X can be written as the topological direct sum of locally compact, σ -compact (hence hemicompact) disjoint { : ∈ } subspaces Xt t T of X. Since Cc(X) is isomorphic to the product t Cc(Xt ) of Fréchet spaces, the space Cc(X) is a Baire space; see, for example, [328]. Gru- enhage and Ma [194] defined and studied the moving off property and proved that if X is locally compact or first-countable, Cc(X) is Baire if and only if X has the moving off property. 34 2 Elementary Facts about Baire and Baire-Type Spaces

Combining results of Lehner [266, Theorem III.2.2, Theorem III.3.1] and Propo- sition 2.15, we note the following proposition.

Proposition 2.18 (Lehner) Cc(X) is Baire-like if and only if for each decreasing sequence (An)n of closed noncompact subsets of X there exists a continuous function f ∈ C(X) that is unbounded on each An.

This yields the useful Proposition 2.19;see[266] and also [225]. Recall that a topological space X is of pointwise countable type if each x ∈ X is contained in a compact set K ⊂ X of countable character in X (i.e., having a countable basis of open neighborhoods). All ﬁrst-countable spaces, as well as Cech-completeˇ spaces (hence locally compact spaces), are of pointwise countable type; see [146], [27].

Proposition 2.19 (Lehner) (i) If X is a locally compact space and Cc(X) is barrelled, then Cc(X) is a Baire-like space. (ii) If X is a space of pointwise countable type and Cc(X) is Baire-like, X is a locally compact space.

Proof (i) Assume Cc(X) is a barrelled space. Then, by Proposition 2.15, the space X is a μ-space. Having in mind Proposition 2.18, we need only to find for a decreasing sequence (An)n of closed, not topologically bounded sets in X a continuous function f ∈ C(X) that is unbounded on each An. By the assumption, on each An there exists fn ∈ C(X) unbounded on An. Note that the proof will be completed if we find a number m ∈ N such that fm is unbounded on each An. Assume that ∈ N ∈ N for each n there exists kn such that fn is bounded on Akn . Since (An)n is decreasing, we may assume that fn is bounded on An+1 for each n ∈ N. Two cases are possible: := (a) A n An is noncompact. Since X is a μ-space, there exists a continuous function f ∈ C(X) that is unbounded on A, and the proof is finished. (b) A is compact. Since X is locally compact, there exists an open set V0 such that A ⊂ V0 whose closure W0 is compact. But f1 is bounded on A2 ∪ W0 and unbounded on A1,soA1 ⊂ A2 ∪ W0. Select

x1 ∈ A1 \ (A2 ∪ W0).

Then there exists an open neighborhood V1, x1 ∈ V1, whose closure W1 is compact, and

V1 ⊂ X \ (A2 ∪ W0).

By a simple induction, we select a sequence (xn)n in X and a pairwise disjoint sequence (Vn)n of open neighborhoods of xn whose closure Wn is compact for each n ∈ N. Since (Vn)n is an open cover of

{xn : n ∈ N}∪A, we deduce that this cover does not admit a ﬁnite subcover. We conclude that {xn : n ∈ N}∪A is not compact. Set L := {xn : n ∈ N}. 2.4 Locally convex spaces with some Baire-type conditions 35

Claim 2.5 L ∪ A is closed in X. Indeed, if x ∈ L \ A, then x ∈ {xk : k>n} for all n ∈ N. Since {xk : k>n}⊂An for each n ∈ N, x ∈ A. This yields that L ∪ A is closed. As every topologically bounded set in X is relatively compact, there exists a continuous function f ∈ C(X) unbounded on L ∪ A. Therefore f is unbounded on each An.

(ii) Fix x ∈ X. By the assumption, there exists a compact set K, x ∈ K, and a decreasing basis (Un)n of open neighborhoods of K. Set Hn := f ∈ C(X) : sup |f(h)|≤n . h∈Un

Then, as is easily seen, the family {Hn : n ∈ N} covers C(X). Since Cc(X) is Baire- like, there exist a compact set D ⊂ X, ε>0, and n ∈ N such that f ∈ C(X) : sup |f(d)| <ε ⊂ Hn. d∈D

Observe that Un ⊂ D. Indeed, if z ∈ Un \ D, we can ﬁnd a continuous function f ∈ C(X) such that f(z)= n + 1 and f(d)= 0 for all d ∈ D. Hence f ∈ Hn,so f(z)≤ n, which provides a contradiction.

We already mentioned that the product of two normed Baire spaces need not be a Baire space. Nevertheless, particular products of Baire spaces are Baire (see the proof of Proposition 2.2 and text below Proposition 2.7). On the other hand, any product of metrizable and separable Baire spaces is a Baire space; see [324]. This fact can be applied to get the following interesting result concerning the products of Baire spaces Cp(X);see[398, Theorem 4.8], [278].

Theorem 2.8(Tkachuk) Let {Cp(Xt ) : t ∈ A} be a family of Baire spaces. Then the product t∈A Cp(Xt ) is a Baire space. = Proof Note that t∈A Cp(Xt ) is isomorphic to Cp(X), where X t∈A Xt .We will need here the following useful fact stating that Cp(X) is a Baire space if and only if TD(Cp(X)) is a Baire space for each countable set D ⊂ X, where as usual TD(f ) := f |D for f ∈ C(X) means the restriction map; see [278, Theorem 3.6]. Let D ⊂ X be a countable set, and set Dt := D ∩ Xt for each t ∈ A. Clearly, WD := {t ∈ A : Dt = ∅} is countable. Since by the assumption each Cp(Xt ) is a Baire space,

TDt (Cp(Xt )) is a Baire space. On the other hand, as is easily seen, = TD(Cp(X)) TDt (Cp(Xt )). t∈WD

As each space TDt (Cp(Xt )) is a metrizable and separable Baire space, the product TD(Cp(X)) is a Baire space. Hence Cp(X) is Baire. 36 2 Elementary Facts about Baire and Baire-Type Spaces

2.5 Strongly realcompact spaces X and spaces Cc(X)

This section deals with the class of strongly realcompact spaces, introduced and studied in [228];seealso[404]. The following well-known characterization of realcompact spaces will be used in the sequel; see [181]or[146].

Proposition 2.20 A completely regular Hausdorff space X is realcompact if and ∗ only if for every element x ∈ X := βX\X there exists hx ∈ C(βX), hx(X) ⊂]0, 1], which is positive on X, and hx(x) = 0. = { −1] ]: ∈ \ } Proof Assume the condition holds. Then X hy 0, 1 y βX X . As each −1] ] −1] ] hy 0, 1 is a realcompact subspace of βX (since hy 0, 1 is homeomorphic to (βX×]0, 1]) ∩ G(hy), where G(hy) means the graph of hy ), then X is also realcompact. Conversely, if X is realcompact and x0 ∈ βX\X = βX\υX, there exists a continuous function f : X → R that cannot be extended continuously to X ∪{x0}. From f(x)= max(f (x), 0) + min(f (x), 0) =

1 + max(f (x), 0) − (1 − min(f (x), 0)) we know that one of the functions g1(x) = 1 + max(f (x), 0),org2(x) = 1 − min(f (x), 0) cannot be extended continuously to X ∪{x0}. Hence there exists a continuous function g : X →[1, ∞[ that cannot be extended continuously to X ∪{x0}.Leth be a continuous extension of the bounded function h := 1/g to βX.Ifh(x0) = 0, we reach a contradiction. Hence h(x0) = 0.

We shall say that X is strongly realcompact [228] if for every sequence (xn)n of elements in X∗ there exists f ∈ C(βX) that is positive on X and vanishes on some subsequence of (xn)n. Clearly, every strongly realcompact space is realcompact. It is known [411, Exercise 1B. 4] that if X is locally compact and σ -compact, then X∗ is a zero set in βX,soX is strongly realcompact. A subset A ⊂ X is said to be C-embedded (C∗-embedded) if every real-valued continuous (bounded and continuous) function on A can be extended to a continuous function on the whole space X. For strongly realcompact spaces, we note the following property. The proof presented below (see [228]) uses some arguments (due to Negrepontis [311]) from [180, Theorem 2.7].

Proposition 2.21 If X is strongly realcompact, every inﬁnite subset D of X∗ contains an inﬁnite subset S that is relatively compact in X∗ and C∗-embedded in βX.

Proof Let (xn)n be an injective sequence in D (i.e., xn = xm if n = m), and let f : βX →[0, 1] be a continuous function that is positive on X and vanishes on a subsequence of (xn)n. Set

−1 −1 S ={xn : n ∈ N}∩f {0},Yn ={x ∈ βX :|f(x)| n },n∈ N, 2.5 Strongly realcompact spaces X and spaces Cc(X) 37 X1 = S ∪ Yn. n

Note that the space X1 is regular and σ -compact; hence it is normal (see ∗ Lemma 6.1). Since S is closed in X1,itisC -embedded in X1. Therefore S is ∗ C -embedded in βX1.AsX ⊂ X1 ⊂ βX, we conclude that βX1 = βX.

This implies that, if X is a strongly realcompact space, every infinite closed subset of X∗ contains a copy of the space βN. On the other hand, Baumgartner and van Douwen [48, Example 1.11] provided a separable first-countable locally compact realcompact space X (hence strongly realcompact by Theorem 2.9 below) for which X∗ contains a discrete countable subset that is not C∗-embedded in βX.This result with [48, Theorem 1.2] can be used to distinguish an example of a locally ∗ compact realcompact space X such that X contains a sequence (xn)n for which there does not exist f ∈ C(βX) that is positive on X and vanishes on (xn)n. The space Q of the rational numbers is not strongly realcompact, but by [139] one gets that Q∗ is a βω-space (i.e., if D is a countable discrete subset of Q∗, and D (the closure in Q∗) is compact, then D = βD). Hence D is C∗-embedded in βQ. It is known (see [181]) that Q∗ contains a countable subset that is not C∗-embedded in βQ. A filter (filterbasis) F on a topological space X is said to be unbounded if there exists a continuous real-valued function f on X that is unbounded on each element of F . We call f unbounded on F . In order to prove Theorem 2.9, we need the following two lemmas.

Lemma 2.4 A filter F on a topological space X is unbounded if and only if there ∈ \ exists x F ∈F F υX, where the closure is taken in βX. := ⊂ Proof Set K F ∈F F , and assume by contradiction that K υX. Then, for each continuous real-valued function f on X there exists an open set Uf ⊂ βX such that K ⊂ Uf and the restriction f |Uf ∩ X is bounded. Note that there exists F ∈ F contained in Uf . Indeed, otherwise the family of sets {F \ Uf : F ∈ F } satisfies the finite intersection property, which leads to a point in K \ Uf .Thisisa contradiction. We proved that there exists F ∈ F contained in Uf . This shows that F is not unbounded. To prove the converse, assume that there exists x ∈ K \ υX. It is known (see = [181]) that υX f ∈C(X) υf (X), where

β υf (X) := {x ∈ βX : f (x) = ∞}.

Then there exists f ∈ C(X) for which the extension f β : βX → R∞ has property ∞ f (x) =∞, where R∞ := R ∪{∞}(the Alexandrov one-point compactiﬁcation). Since x ∈ F for each F ∈ F ,themapf is unbounded on F .

Lemma 2.5 Each unbounded ﬁlterbasis F on a topological space X is contained in an unbounded ultraﬁlter U on X. 38 2 Elementary Facts about Baire and Baire-Type Spaces

Proof If M := {M ⊂ X :∃F ∈ F; F ⊂ M}, then M is an unbounded filter on X. ∈ \ A By Lemma 2.4, there exists x F ∈M F υX. Let be the family of all filters G M ∈ A on X containing and such that x F ∈G F. Order by inclusion. Since there exists a maximal chain in A , its union U is an ultrafilter on X containing F ∈ U such that x F ∈U F , and we use Lemma 2.4 to conclude that is unbounded on X.

We are ready to prove the following characterization of strongly realcompact spaces. Parts (i) and (ii) were proved in [228]; part (iii) is from [404].

Theorem 2.9 (i) A topological space X is strongly realcompact if and only if X is realcompact and X∗ is countably compact. Hence every locally compact realcompact space is strongly realcompact. (ii) Every strongly realcompact space of pointwise countable type is locally compact. (iii) A realcompact space X is strongly realcompact if and only if for each sequence (Fn)n of unbounded ﬁlters (ﬁlterbases) on X there exists a continuous F real-valued function on X and a subsequence ( nk )k such that f is unbounded F on each nk .

Proof (i) Suppose X is strongly realcompact. Let P ⊂ X∗ be an inﬁnite set, and let (xn)n be an injective sequence in P . There exists a continuous function f : βX → [ ] 0, 1 that is positive on X and zero on some subsequence (xkn )n of (xn)n. Then

{ : ∈ N}⊂ −1 ⊂ ∗ xkn n f (0) X .

Hence { : ∈ N}d ⊂ −1 xkn n f (0). { : ∈ N}d d Note that xkn n is nonempty, where A is the set of all accumulation points of a set A. This proves that P has an accumulation point in X∗. To prove the converse, assume that X is realcompact and every infinite subset of X∗ has an accumulation point in X∗. ∗ Let (xn)n be a sequence in X .IfP ={xn : n ∈ N} is finite, then (since X is realcompact) there exists a continuous function f : βX →[0, 1] that is positive on X and zero on a subsequence of (xn)n.IfP ={xn : n ∈ N} is infinite, take p ∈ P d \ X. Then there exists a continuous function f : βX →[0, 1] that is positive on X and vanishes on p. Note that for every r>0thesetP ∩ f −1([0,r)) is infinite since f −1([0,r))is a neighborhood of the point p ∈ P d . We consider two possible cases: Case 1. The set P ∩ f −1(0) is infinite. Then f is positive on X and zero on some subsequence of the sequence (xn)n. Case 2. The set P ∩ f −1(0) is finite. As for every r>0thesetP ∩ f −1([0,r))is infinite, there exists an injective sequence (tn)n in P such that the sequence (f (tn))n 2.5 Strongly realcompact spaces X and spaces Cc(X) 39 is strictly decreasing and converges to zero. Set P0 ={tn : n ∈ N}, s0 = 1 and sk ∈ (f (tk+1), f (tk)) for all k ∈ N. Then (sk)k is decreasing and converges to zero. Set

−1 Fk = f ([sk,sk−1]) for k ∈ N. Then Fk are compact and tk ∈ Fn if and only if k = n. Moreover, −1 X ⊂ f ((0, 1]) = Fk k and

P0 ∩ Fk ={tk},k∈ N. = ∈ −1 −1 ] → ∈ d If f(x) c>0, then x f ((2 c,1 ). Since f(tk) 0, we have that x/P0 . ∈ d ∈ −1 ∈ Hence, if x P0 , then x f (0). Hence x/ k Fk. We showed that d ∩ =∅ P0 Fk . k

Since X is realcompact, for every k ∈ N there exists a continuous function fk : βX →[0, 1] that is positive on X and zero on tk.Nextset k = −1 [ −1 ] ∈ N Tn fk ( n , 1 ), n, k . Then ⊂ −1 ] = k X fk ((0, 1 ) Tn n ∈ k ∈ N and tk / Tn for all k,n . Moreover, ⊂ ∩ ⊂ ∩ k ∩ ∩ k ⊂ ∩ ={ } X Fk X Fk Tn ,P0 (Fk Tn ) P0 Fk tk . k k n

∈ ∩ k As tk / Fk Tn , ∩ ∩ k =∅ P0 (Fk Tn ) for all n, k ∈ N. Hence P0 ∩ W =∅, and d ∩ ⊂ d ∩ =∅ P0 W P0 Fk , k = ∩ k ∩ =∅ where W k n Fk Tn . Therefore P0 W . We showed that there exist an inﬁnite subset P0 of P and an inﬁnite sequence of compact sets (Kn)n such that X ⊂ Kn ⊂ βX, Kn ∩ P0 =∅. n n 40 2 Elementary Facts about Baire and Baire-Type Spaces

For every n ∈ N,letgn : βX →[0, 1] be a continuous function such that

gn|Kn = 1,gn|P0 = 0. = −n : →[ ] Put g n 2 gn. The function g βX 0, 1 is continuous, positive on X, and zero on some subsequence of the sequence (xn)n. This shows that for every ∗ sequence (xn)n in X there exists a continuous function on βX that is positive on X and vanishes on some subsequence of (xn)n. (ii) Assume X is a strongly realcompact space of pointwise countable type and is not locally compact. Then there exist x0 ∈ X for which there does not exist a relatively compact open neighborhood, and a compact set K with x0 ∈ K that admits a countable (decreasing) basis (Un)n of neighborhoods of K. For every n ∈ N, choose xn ∈ (Un \ X), where the closure is taken in βX. Note that

d (βX \ K)∩{xn} =∅.

Indeed, let x ∈ (βX \ K).LetV ⊂ βX be an open neighborhood of K such that ∈ \ ∈ N ⊂ ∩ ⊂ x (βX V). Then there exists n0 such that Un0 V X,soUn0 V . Since { }d ⊂ ⊂ xn Un0 V, d d x ∈ βX \{xn} . Hence {xn} ⊂ K. This shows that X is not strongly realcompact, a contradiction. (iii) Assume X is strongly realcompact and each Fn is an unbounded ﬁlterbasis on X. For each n ∈ N, there exists an accumulation point of Fn,sayxn ∈ βX\ υX. Since X is a realcompact space, υX = X.AsX is strongly realcompact, there exists ∈ a subsequence (xnk )k of (xn)n, and a positive continuous function g C(X) such ≤ β = ∈ N := −1 ∈ that g(x) 1 and g (xnk ) 0 for all k . Then f g C(X). Hence f is F unbounded on each nk since ∈ \ xnk F υf (X). ∈F F nk This proves one direction of the claim (iii). To prove the converse, assume that (xn)n is a sequence in βX\ X. Then, for each n ∈ N there exists a ﬁlter Fn on X that converges to xn in the space βX.But xn ∈ F, xn ∈/ X = υX. F ∈Fn

This shows that each Fn is unbounded on X. By the assumption, there exists a F F ∈ F subsequence ( nk )k of ( )n and f C(X) that is unbounded on each nk . Set − g(x) := (1 +|f(x)|) 1 for each x ∈ X. Clearly, the function g is positive on X and is continuous and bounded. Therefore there exists a continuous extension gβ of g to βX, and clearly β = g (xnk ) 0. This proves that X is strongly realcompact. 2.5 Strongly realcompact spaces X and spaces Cc(X) 41

Example 2.1 There is a strongly realcompact space that is not locally compact. The space RN is realcompact and not strongly realcompact.

Proof Let P be a countable and nonempty subset of N∗. Note that the subspace X := N ∪ P of βN is a Lindelöf space. Hence it is a realcompact space. On the other hand, since every countable and closed subset of βN is ﬁnite (see [411, p. 71]) one gets that the space X∗ = N∗ \ P is countably compact. Now Theorem 2.9 is applied to deduce that X is strongly realcompact. Note that X is not locally compact. The second statement follows directly from Theorem 2.9.

If D is an absolutely convex subset of Cc(X),ahold K of D is a compact subset of βXsuch that f ∈ C(X) belongs to D if its continuous extension f β : βX → βR is identically zero on a neighborhood of K. The intersection k(D) of all holds of an absolutely convex set D in Cc(X) is again a hold. k(D) is called a support of D.If, moreover, D is bornivorous, k(D) is contained in υX;see[328, Lemma 10.1.9]. We also need the following fact due to Valdivia [419];seealso[328].

Lemma 2.6 (Valdivia) Let E be a barrelled space. Let (An)n be an increasing sequence of absolutely convex closed subsets of E covering E. Then, for every bounded set B ⊂ E there exists m ∈ N such that B ⊂ mAm.

Proof Assume that for each n ∈ N there exists xn ∈ B \ nAn. Then the sequence −1 (yn)n, yn := n xn converges to zero in E, and yn ∈/ An for all n ∈ N.Let(Un)n be a decreasing sequence of absolutely convex neighborhoods of zero in E such that Un+1 + Un+1 ⊂ Un and yn ∈/ An + Un for all n ∈ N. Then

yn ∈/ An + Un+1 for all n ∈ N. Set U := (An + Un). n

The set U is closed, absolutely convex and absorbing in E and yn ∈/ U for all n ∈ N. Hence U is a barrel in E. Since E is barrelled, U is a neighborhood of zero in E. This proves that U contains almost all elements of the sequence (yn)n,a contradiction.

We are prepared to prove the following result from [228].

Theorem 2.10 (KakolÐ ˛ Sliwa)« (i) If X is a strongly realcompact space, Cc(X) is Baire-like and bornological. (ii) If X is locally compact, Cc(X) is Baire-like and bornological if and only if X is realcompact. (iii) If X is a space of pointwise countable type, Cc(X) is bornological and Baire- like if and only if X is strongly realcompact. 42 2 Elementary Facts about Baire and Baire-Type Spaces

Proof (i) Let (Dn)n be a bornivorous sequence in Cc(X) (i.e., every bounded set in Cc(X) is absorbed by some Dn). We may assume that every bounded set in Cc(X) is ⊂ ∈ N contained in some Dn. Note that k(Dm0 ) X for some m0 . Indeed, otherwise for every n ∈ N there exists xn ∈ k(Dn) \ X. Since X is strongly realcompact, there exists a continuous function f : βX →[0, 1] that is positive on X and zero on some subsequence of (xn)n. Since (Dn)n is increasing, we may assume that f(xn) = 0, n ∈ N.Thesets −1 Am ={y ∈ βX: f(y)> m } are open in βXand compose an increasing sequence covering X. Since xn ∈/ An for n ∈ N, where the closure is taken in βX,wehavek(Dn) ⊂ An for every n ∈ N.This implies that An is not a hold of Dn for any n ∈ N. Hence there exists a sequence

fn ∈ Cc(X) \ Dn

β such that the extension fn = 0 on some neighborhood of An. As the sequence (fn)n converges to zero in Cc(X), there exists p ∈ N such that fn ∈ Dp for all n ∈ N,a contradiction. ∈ N ⊂ We proved that there exists m0 such that k(Dm0 ) X. Next we show that there exist m m0 and r>0 such that f ∈ C(X) : sup |f(x)| 0 and n m0 such that f ∈ C(X) : sup |f(x)|

fn ∈ C(X) \ Dn

−1 such that |fn(x)| 0, and a compact subset S of X such that f ∈ C(X) : sup |f(y)| <ε ⊂ Wn. y∈S

Hence Un ⊂ S. We proved that X is locally compact. Theorem 2.9 is applied to deduce that X is strongly realcompact. For the converse, we again apply Theorem 2.9 and the previous case.

From Theorem 2.9, we know that a realcompact space X for which βX \ X is countably compact is strongly realcompact. As concerns the converse to Theo- rem 2.10, we note only the following fact.

Proposition 2.22 If Cc(X) is a Baire space and X is realcompact, then βX \ X is pseudocompact (i.e., its image under any real-valued continuous function is bounded).

Proof Assume that X∗ := βX\ X is not pseudocompact. Then there exists a locally ∗ ﬁnite sequence (Un)n of open disjoint subsets in X . Then, by regularity, we obtain a sequence (Vn)n of open nonempty sets in βX such that

∗ ∗ ∅=Vn ∩ X ⊂ Vn ∩ X ⊂ Un, where the closure is taken in βX. Since X is realcompact, An = Vn, where An := ∗ Vn \ X = Vn ∩ X. As every topologically bounded set in X is relatively compact by Proposition 2.15 (since Cc(X) is barrelled), An is not topologically bounded in X for n ∈ N. Since Cp(X) is a Baire space, by [266] there exists a continuous function f ∈ C(X) and a subsequence (An )k such that f |An is unbounded for each k ∈ N. k k ∞ Let R∞ be the Alexandrov one-point compactiﬁcation of R, and let f : βX → R∞ be the continuous extension of f . As each f |An is unbounded, there exists a ∞ k sequence (xk)k such that f (xk) =∞for each k ∈ N and

x ∈ A \ X ⊂ U . k nk nk ∗ Since the sequence (Un )k is locally ﬁnite in X , we deduce that the sequence (xk)k k ∞ ∞ has an adherent point x ∈ X.Butf (xk) =∞for each k ∈ N,sof(x)= f (x) = ∞. This provides a contradiction since f(X)⊂ R. We proved that βX \ X is pseudocompact.

We need the following fact following from [285, Theorem 5.3.5]. 44 2 Elementary Facts about Baire and Baire-Type Spaces

Lemma 2.7 If there exists an inﬁnite family K of nonempty compact subsets of X such that (i) for every compact set L of X there exists K ∈ K with K ∩ L =∅and (ii) any inﬁnite subfamily of K is not discrete, then Cc(X) is not a Baire space.

There are several examples of barrelled spaces Cc(X) that are not Baire; see, for example, [266], [398], [278]. The next example, from [228], is motivated by [48, Example 1.11].

Example 2.2 (K˛akolÐSliwa« ) There exists a locally compact and strongly realcompact space X such that the bornological Baire-like space Cc(X) is not Baire.

Proof Let X be the set R of reals endowed with a topology deﬁned as follows: (a) For every t ∈ Q,theset{t} is open in X. (b) For every t ∈ R \ Q, there exists a sequence (tn)n ⊂ Q that converges to t such that the sets

Vn(t) ={t}∪{tm : m n}, n, m ∈ N, form a base of neighborhoods of t in X. (c) For all dense sets A,B ⊂ Q in the natural topology of R,thesetA ∩ B is non-empty, where the closure is taken in X. As the space X is a locally compact and realcompact space, it is strongly realcompact by Theorem 2.9. We prove that X satisﬁes conditions (i) and (ii) from Lemma 2.7.Let(Pn)n be a sequence of pairwise disjoint ﬁnite subsets of R \ Q such that for any subsequence R (Pnk )k the set k Pnk is dense in . Set ={ n : } Pn tk 1 k mn for all n ∈ N. For all n, m ∈ N,theset = n Kn,m Vm(tk ) 1kmn is nonempty and compact in X.

Claim 2.6 K ={Kn,m : n, m ∈ N} satisﬁes (i). Indeed, any compact subset L of X is contained in a set of the form

1 m V1(t ) ∪···∪V1(t ) ∪{p1,...,pk},

1 2 m where m, k ∈ N and t ,t ,...,t ∈ R \ Q, p1,...,pk ∈ Q. In fact, the set L ∩ (R \ Q) is ﬁnite since the open cover

{V1(t) : t ∈ L ∩ (R \ Q)}∪{{q}:q ∈ L ∩ Q} of L has a finite subcover. On the other hand, L \ {V1(t) : t ∈ L ∩ (R \ Q)} 2.5 Strongly realcompact spaces X and spaces Cc(X) 45 is finite since Xd ⊂ (R \ Q). Hence K indeed satisfies (i).

Claim 2.7 K satisﬁes (ii).

n Indeed, otherwise some inﬁnite subfamily {K : n ∈ N} of K is discrete. But = 2n = 2n+1 then sets A n K and B n K are disjoint and closed in X. Applying the property of (Pn)n, one gets that A ∩ Q and B ∩ Q are dense in R. Hence, by (c), the set (A ∩ Q) ∩ (B ∩ Q) is nonempty, a contradiction. We proved that Cc(X) is not Baire.

Tkachuk [398] provided an example of a countable space X that has exactly one nonisolated point, has no inﬁnite topologically bounded sets (hence Cp(X) is barrelled by Proposition 2.17) and Cp(X) is not Baire. We show another example of this type. Set X := N2 ∪{x} for x/∈ N2, where all points of N2 are isolated in X, and set Xn := {(m, n) : m ∈ N} for each n ∈ N. The basis B(x) at x is formed by the sets

{U ⊂ X : x ∈ U, |{n :|(X \ U)∩ Xn|=∞}|< ∞}.

The space X (originally deﬁned by Arens) is completely regular and Hausdorff. Since X is countable, it is Lindelöf.

Example 2.3 The space Cp(X) for the Arens space X is barrelled and not Baire.

Proof If K ⊂ X is compact, K is ﬁnite (so X is hemicompact). Indeed, note that |K ∩ Xn| < ∞ for each n ∈ N and

|{n ∈ N : K ∩ Xn = ∅}| < ∞.

Clearly, Cp(X) is metrizable since X is countable. By Proposition 2.17 and Propo- sition 2.15, the space Cc(X) = Cp(X) is barrelled. We show that Cp(X) is not Baire. First observe that for each f ∈ C(X) there exists n ∈ N such that the restriction f |Xn is bounded. Indeed, otherwise, if there exists f ∈ C(X) that is unbounded on each Xn, then f is unbounded on each open neighborhood of the point x. This implies that f is discontinuous at x, a contradiction. Now set Bm,n := f ∈ C(X) : sup |f(y)|≤n y∈Xm ∈ N = for each n, m .ThesetsBm,n are absolutely convex and closed. Clearly, C(X) ∈ N n Bm,n. Assume that Cp(X) is a Baire space. Then there exist m, n such that Bm,n is a neighborhood of zero in Cp(X). Hence there exists a compact set K ⊂ X and ε>0 such that f ∈ C(X) : sup |f(y)| <ε ⊂ Bm,n. (2.7) y∈K 46 2 Elementary Facts about Baire and Baire-Type Spaces

We claim that Xm ⊂ K (which will provide a contradiction). If there exists y ∈ Xm \ K, then there exists f ∈ C(X) such that f(z)= 0 for each z ∈ K and f(y)>n.This yields a contradiction with (2.7).

The following problem is motivated by the previous results.

Problem 2.1 Characterize a strongly realcompact space X in terms of topological properties of Cc(X) (or Cp(X)).

2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X)

In this section, we characterize pseudocompact spaces. For example, we show that X is pseudocompact, a Warner-bounded set or Cp(X) is a (df )-space if and only if for each sequence (μn)n in the dual Cc(X) of Cc(X) there exists a sequence (tn)n ⊂ (0, 1] such that (tnμn)n is weakly bounded, strongly bounded, or equicontinuous, respectively. This result will be used to provide an example of a (df )-space Cc(X) that is not a (DF )-space. This solves an open question; see [231]. Parts of this section will be used to present concrete examples of quasi-Suslin spaces that are not K-analytic. Buchwalter [68] called a topological space X Warner bounded if for every sequence (Un)n of nonempty open subsets of X there exists a compact set K ⊂ X such that Un ∩ K = ∅ for inﬁnitely many n ∈ N. In fact, Buchwalter required that (Un)n be disjoint, but due to regularity of X this condition could be omitted. Warner has already observed [413] that any Warner-bounded space is pseudocompact. First we provide the following useful analytic characterization of Warner boundedness; see [232]. In this section, X always means a completely regular Hausdorff topological space.

Theorem 2.11 (KakolÐSaxonÐTodd) ˛ Cc(X) does not contain a dense subspace RN if and only if X is Warner bounded.

For the proof, we need the following lemma.

Lemma 2.8 (a) An lcs E contains a dense subspace G of RN if and only if there exists a sequence (wn)n of nonzero elements in E such that every continuous seminorm in E vanishes at wn for almost all n ∈ N. (b) If an lcs E contains a dense subspace G of RN, then the strong dual (E,β(E,E))contains the space ϕ.

Proof (a) Assume that E contains a sequence as mentioned. For (wn)n there exists a biorthogonal sequence (vn,un)n in F × F , where (vn)n is a subsequence of 2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X) 47

(wn)n, F is a linear span of vn, n ∈ N, and F is spanned by un, n ∈ N. Note that F is isomorphic to the linear span G of the unit vectors of RN. Now assume that N R contains a dense subspace G.Let(pn)n be a fundamental sequence of contin- := n −1 uous seminorms on G. Then each Gn i=1 pi (0) is an inﬁnite-dimensional subspace of G. Therefore we can ﬁnd wn ∈ Gn \{0} for each n ∈ N, as required. (b) It is well known that the strong dual of RN is the space ϕ; see, for example, [328]. On the other hand, it is also well known that every bounded set in the completion of Gˆ = RN is contained in the Gˆ -completion of a bounded set in G; see also [328, Observation 8.3.23]. This completes the proof of (b).

Now we are ready to prove Theorem 2.11.

N Proof Assume that Cc(X) contains a dense subspace of R . Then there exists a sequence (fn)n of nonzero elements of C(X) that vanishes for almost all n ∈ N on any compact subset of X. Then, for each n ∈ N there exists an open nonzero set Un in X such that fn(y) = 0 for all y ∈ Un. Therefore every compact set K in X misses Un for almost all n ∈ N. This shows that X is not a Warner-bounded set. To prove the converse, assume that (Un)n is a sequence of nonempty open sets in X such that almost all of them miss each compact set in X. Then we can select a sequence (fn)n of nonzero continuous functions on X such that each fn(X \ Un) = {0}. Hence each continuous seminorm on Cc(X) vanishes on almost all elements of the sequence (fn)n. Now we apply Lemma 2.8.

Theorem 2.12 (see [231]) looks much more interesting if we have already in mind Theorem 2.11.

Theorem 2.12 (KakolÐSaxonÐTodd) ˛ The following assertions are equivalent for X: (i) X is pseudocompact. ∗ (ii) For each sequence (μn)n in the weak dual F of Cc(X), there exists a sequence (tn)n ⊂ (0, 1] such that (tnμn)n is bounded in F . ∗ (iii) The weak dual F of Cc(X) is docile (i.e., every inﬁnite-dimensional subspace of F contains an inﬁnite-dimensional bounded set). N (iv) Cc(X) does not contain a copy of R .

Proof (i) ⇒ (ii): Take a sequence (μn)n in F . Set −1 tn := (μn+1) for each n ∈ N. Since X is pseudocompact, (tnμn(f ))n is bounded for each f ∈ C(X). (ii) ⇒ (iii) holds for any lcs. (iii) ⇒ (i)isclear. N (iii) ⇒ (iv): Assume G is a subspace of Cc(X) isomorphic to R . It is clear that the weak∗ dual of RN, and hence (G,σ(G,G)), is not docile. Using the HahnÐ Banach theorem, we extend elements of G to the whole space C(X) that generate ∗ a nondocile subspace of the weak dual of Cc(X). This contradicts (iii). 48 2 Elementary Facts about Baire and Baire-Type Spaces

(iv) ⇒ (i): Suppose that X is not pseudocompact. Then there exists a sequence (Un)n of disjoint open nonempty sets in X that is locally finite. Choose xn ∈ Un and ∈ = \ ={ } ∈ N fn C(X) such that fn(xn) 1 and fn(X Un) 0for each n . Note that, because the sequence (Un)n is locally finite, the series n anfn converges in Cc(X) N N for any scalar sequence (an)n in R . Define a map T : R → Cc(X) by T ((an)n) := anfn. n

The map T is injective. Indeed, since for each evaluation map δxm we have ∈ δxm (Cc(X), σ (Cc(X) ,Cc(X))), and = δxm anfn am, n then indeed T is an injective open map. We show that T is continuous. Each partial sum map TN deﬁned by N TN ((an)n) := anfn n=1 is continuous, and T is the pointwise limit of the sequence (TN )N . Since the space N R is Baire, the sequence (TN )N is equicontinuous, and T is continuous by the classical BanachÐSteinhaus theorem. Consequently, the space Cc(X) contains a copy of RN.

Many barrelled spaces Cc(X) are not Baire-like; for example, Cc(Q) is not Baire- like (although barrelled by Proposition 2.15) since Q is not locally compact, and we apply Proposition 2.19. By Theorem 2.5, the space Cc(Q) contains ϕ. It is interesting that owing to Theorem 2.12 each space Cc(X) that contains the nondocile space ϕ contains also the docile space RN. So we deduce that every barrelled non-Baire- N N like space Cc(X) contains both spaces ϕ and R . Note that Cc(R) contains R but not ϕ. Theorem 2.11 and Lemma 2.8 apply to simplify essentially Warner’s fundamental theorem [413, Theorem 11].

Theorem 2.13 The following assertions are equivalent: (i) X is Warner bounded. [ ]:={ ∈ : | |≤ } (ii) X, 1 f C(X) supx∈X f(x) 1 absorbs bounded sets in Cc(X). (iii) Cc(X) has a fundamental sequence of bounded sets. (iv) Every Cauchy sequence in Cc(X) is a Cauchy sequence in the space Cb(X) of the continuous bounded functions on X with the uniform Banach topology. (v) X is pseudocompact, and Cc(X) is sequentially complete. (vi) X is pseudocompact, and Cc(X) is locally complete. 2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X) 49

Proof (i) ⇒ (ii): Let B ⊂ Cc(X) be a bounded set. Hence B is uniformly bounded on each compact set in X. Assume B is not absorbed by [X, 1]. Then for each n ∈ N there exist fn ∈ B, and xn ∈ X such that |fn(xn)| >n. Consequently, for each n ∈ N there exists an open neighborhood Un of xn such that |fn(x)| >nfor all x ∈ Un. Since X is Warner bounded, there exists a compact set K ⊂ X such that K ∩ Un = ∅ for almost all n ∈ N, a contradiction since {fn : n ∈ N}⊂B must be uniformly bounded on K. (ii) ⇒ (iii): The sets n[X, 1] for n ∈ N form a fundamental sequence of bounded sets. (iii) ⇒ (i): By the assumption, the strong dual of Cc(X) is metrizable. Now we apply Lemma 2.8 (b) and Theorem 2.11 to complete the proof. (i) ⇒ (iv): By (i) ⇒ (ii), the space X is pseudocompact. To prove the second part, let (fn)n be a null sequence in Cc(X). We show that (fn)n is a null sequence in the uniform Banach topology of Cb(X). Assume this fails. Then there exist a subsequence (hn)n of (fn)n, ε>0, and a sequence (Un)n of nonzero open sets in X such that |hn(x)| >ε for all x ∈ Un. Since (hn)n converges to zero uniformly on compact sets of X,theymissUn for almost all n ∈ N. This contradicts (i). (iv) ⇒ (v): By (iv), X is pseudocompact and Cb(X) is complete. (v) ⇒ (vi): Any sequentially complete lcs is locally complete. (vi) ⇒ (ii): In a locally complete lcs, barrels absorb bounded sets; this fact is elementary; see, for example, [328, Corollary 5.1.10]. Therefore the set [X, 1] (which is clearly closed, absolutely convex and absorbing) absorbs bounded sets.

It is worth noticing here another interesting characterization of a completely regular Hausdorff space X to be pseudocompact (see [416]): X is pseudocompact if and only if every uniformly bounded pointwise compact set H in the space Cb(X) is weakly compact. In order to prove the main result of this section, we shall need two additional lemmas.

Lemma 2.9 If every countable subset of X is relatively compact, [X, 1] is bornivorous in Cc(X).

Proof If a bounded set A ⊂ Cc(X) is not absorbed by [X, 1], there exist two sequences, (xn)n in X and (fn)n in A, such that (fn(xn))n is not bounded. Since the closure of the set {xn : n ∈ N} is compact in X, [K,1] is a neighborhood of zero in Cc(X) that does not absorb A, a contradiction.

We also need the following useful fact; see [328] and [360].

1 Proposition 2.23 An lcs E is locally complete if and only if it is -complete (i.e., ∈ 1 for each (tn)n and each bounded sequence (xn)n in E, the series n tnxn converges in E). 50 2 Elementary Facts about Baire and Baire-Type Spaces

Proof Let ξ be the original topology of E. Assume that E is locally complete, 1 and ﬁx arbitrary (tn)n in and a bounded sequence (xn)n in E. Then the closed, absolutely convex hull B of (xn)n is a Banach disc. Since for r>swe have tnxn − tnxn ∈ (|ts|+···+|tr |)B, n≤r n~~k,wehave −k −r −k+1 tnxn − tnxn ∈ (2 +···+2 ) ⊂ 2 B. n≤r n~~

An lcs E is called dual locally complete [424]if(E,σ(E,E)) is locally complete. The space E is called dual 1-complete if (E,σ(E,E))is 1-complete. Note also a dual version of Proposition 2.23;see[360].

~~Proposition 2.24 For an lcs E, the following assertions are equivalent: (i) E is dual locally complete. 2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X) 51~~

~~(ii) E is dual 1-complete. (iii) If A is an absorbing and absolutely convex subset of E, f is a linear functional over E and f |A is continuous on A, then f is continuous on E.~~

Proof Because of Proposition 2.23, it is enough to prove (ii) ⇒ (iii) and (iii) ⇒ (ii). (ii) ⇒ (iii): Let A be an absolutely convex and absorbing closed subset of E, and let f be a linear functional on E that is continuous on A. We may assume that there exists x ∈ E such that f(x)= 1. For each n ∈ N,setAn := nA and

~~Cn := nA ∩{y ∈ E : f(y)= 0}.~~

Then, for each n ∈ N there exists a continuous linear functional fn with fn(x) = −n n 1 and |fn(y)| < 2 for each y ∈ Cn. Clearly, the sequence (2 (fn+1 − fn))n is σ(E,E)-bounded, so by assumption we deduce that −n n f = f1 + 2 [2 (fn+1 − fn)] n is continuous. 1 (iii) ⇒ (ii): Choose an arbitrary sequence (tn)n in , and a sequence (fn)n in E that is σ(E,E)-bounded. Set f(x)= tnfn(x) n for each x ∈ E. We need to show that f is continuous. Because of (iii), it is enough ◦ to show that f |A is continuous at zero, where A := {fn : n ∈ N} .Fixε>0, and ∈ N | | −1 choose k with n>k tn < 2 ε. Select a neighborhood of zero U in E such that −1 tnfn (x) < 2 ε n≤k for all x ∈ U. Finally, let x ∈ A ∩ U. Then |f(x)|≤ tnfn (x) + |tn| <ε. n≤k n>k The proof is completed.

A subset K of X is the support of some continuous linear functional λ on Cc(X) (denoted by supp λ) if it is the intersection of closed sets M ⊂ X having the property that λ(f ) = 0iff ∈ C(X) and f(M)={0}. We need the following lemma; see [231].

Lemma 2.10 With E := Cc(X), if one of (E ,β(E ,E)), (E ,σ(E ,E)), or (E,n(E,E))is both docile and locally complete, then [X, 1] is bornivorous, where (E,n(E,E)) denotes the dual of E endowed with the topology generated by the norm λ:=sup{|λ(f )|:f ∈[X.1]}. 52 2 Elementary Facts about Baire and Baire-Type Spaces

Proof Assume [X, 1] is not bornivorous. Then there exist a sequence (xn)n in X and a bounded sequence (fn)n in Cc(X) such that fn(xn)>nfor each n ∈ N. Since the sequence (fn)n is bounded, the set {xn : n ∈ N} is infinite. We may assume that the sequence (xn)n is injective (i.e., xn = xm for n = m). Hence the evaluation func- ⊂ = tionals δxn form a Hamel basis for their linear span D E . Clearly, γ n bnδxn for γ ∈ D, where all but finite numbers bn are zero. Then supp γ := {xn : bn = 0} is finite, so there exists

~~min γ := min{|bn|:xn ∈ supp γ }.~~

By the assumption about docility, we deduce that there exists a linearly independent sequence (γn)n in D that is bounded in each dual mentioned above. By the local completeness, the series n anγn converges for each absolutely summable sequence (an)n (by Proposition 2.23), and n anγn converges pointwisely for each f ∈[X, 1]. Choose inductively an absolutely summable sequence (an)n of nonzero elements such that an → 0 and |akγk| < |an| min γn k>n ∈ N := ∈ for each n . Hence γ n anγn E . This implies that there exists a compact set K ⊂ X such that γ is bounded on the neighborhood of zero [K,1] in Cc(X). On the other hand, linear independence of (γn)n yields that n supp γn is an inﬁnite subset of (xn)n. Then, since (fn)n is uniformly bounded, we deduce that n supp γn is not a subset of K.Letp ∈ N such that supp γp is not a subset of K.Fixy ∈ supp γp \ K.Theset

A := K ∪{x = y : x ∈ supp γp} is closed and misses y, and there exists g ∈[X, 1] with g(y) = 1 and g(A) ={0}. Since cg ∈[K,1] for each scalar c, we deduce that linear functional γ vanishes on g. On the other hand, we note that |γ(g)|= anγn(g) = anγn(g) n n≤p ≥|apγp(g)|− |akγk(g)|≥|ap| min γp − |ak|γk > 0. k>p k>p This provides a contradiction.

~~Buchwalter and Schmets [69, Theorem 4.1] proved the following.~~

∞ Proposition 2.25 Cc(X) is -barrelled if and only if the weak dual of Cc(X) is locally complete if and only if a countable union of support sets in X is relatively compact provided it is topologically bounded. 2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X) 53

An lcs E is called a (DF )-space if E admits a fundamental sequence of bounded sets and is ℵ0-quasibarrelled (i.e., any countable union of equicontinuous sets that is β(E ,E)-bounded is equicontinuous). An lcs E is ℵ0-barrelled if any countable union of equicontinuous sets in E that is σ(E,E)-bounded is equicontinuous. By Warner ([413];seealso[328, Theorem 10.1.22]) we have the following proposition.

Proposition 2.26 Cc(X) is a (DF )-space if and only if each countable union of compact sets in X is relatively compact if and only if Cc(X) has a fundamental sequence of bounded sets and is ℵ0-barrelled.

Recall that an lcs E is called a (df )-space if it admits a fundamental sequence of bounded sets and is c0-quasibarrelled (i.e., every null sequence in (E ,β(E ,E)) is equicontinuous). The classes of (DF )-spaces and (df )-spaces will be discussed in the following chapters. Clearly, every (DF )-space is a (df )-space. It was for a long time unknown whether every (df )-space is (DF ).K˛akol, Saxon and Todd [231] provided an example of a (df )-space Cc(X) that is not a (DF )-space; see Example 2.4 below. To present Example 2.4, first we recall some additional concepts and facts. An lcs E satisfies the countable neighborhood property (cnp) if for every sequence (Un)n of neighborhoods of zero in E there exists a sequence (an)n of positive scalars such that n anUn is a neighborhood of zero. From Warner [413], the space Cc(X) is a (DF )-space if and only if Cc(X) satisfies the cnp. It is known that all support sets satisfy the cnp; see [213], [286]. We also need a couple of definitions concerning the Borel measures;see[173]. A Borel measure μ on a space X is a σ -additive real-valued finite function on all Borel subsets of X. It is called regular if its negative and positive parts satisfy

~~μ(A) = sup {μ(K) : K ⊂ A is compact}.~~

It is known that for each regular Borel measure there exists a smallest closed set M in X such that μ vanishes on each Borel set that misses M, and M is called the support of μ (again we use the notation supp μ), possibly noncompact. Nevertheless, one has supp λ = supp μ if either μ, a nonnegative regular Borel measure, or λ ∈ Cc(X) is given and the other is appropriately chosen. We are prepared to formulate and prove the main result of this section; see [231]. Note that the equivalence (ii) ⇔ (xi) has already been proved by Mazon [283]. Also, (iv) ⇔ (viii) was known to McCoy and Todd [286]. By Cd (X) we denote the space C(X) endowed with the topology of the uniform convergence on support sets of X.

Theorem 2.14 (KakolÐSaxonÐTodd) ˛ The following assertions are equivalent for E := Cc(X): (i) E is a (df )-space. (ii) (E,β(E,E))is a Fréchet space. (iii) (E,β(E,E))is a Banach space and equals (E,n(E,E)). (iv) (E,n(E,E))is a Banach space. (v) (E,β(E,E))is docile and locally complete. 54 2 Elementary Facts about Baire and Baire-Type Spaces

(vi) (E,σ(E,E))is docile and locally complete. (vii) X is pseudocompact, and (E,σ(E,E))is locally complete. (viii) Every countable union of support sets in X is relatively compact. (ix) For each sequence (μn)n in E , there exists a sequence (tn)n ⊂ (0, 1] such that {tnμn)n is equicontinuous. (x) Cd (X) satisﬁes the cnp. (xi) Each regular Borel measure on X has compact support.

Proof (i) ⇒ (ii): It is clear that (E,β(E,E))is locally complete (since the strong dual of a c0-quasibarrelled space is locally complete [328, Proposition 8.2.23(b)]). But then (E,β(E,E)) is a Fréchet space since E has a fundamental sequence of bounded sets. Conditions from (ii) to (vi) are equivalent since any of them implies that [X, 1] is bornivorous owing to Lemma 2.10. Then (E,β(E,E)) = (E,n(E, E)), so applying the BanachÐSteinhaus theorem, we deduce that σ(E,E)-bounded sets are β(E,E)-bounded. (vi) ⇔ (vii): This follows from Theorem 2.12. (vii) ⇒ (viii): Since X is pseudocompact, every subset of X is topologically bounded. Since by assumption (E,σ(E,E)) is locally complete, applying Propo- sition 2.25 we have that (viii) holds. (viii) ⇒ (i): Clearly, every singleton subset of X is a support set, so by Lemma 2.9 we note that {[X, n]:n ∈ N} is a fundamental sequence of bounded ∞ sets in E. Hence, by Proposition 2.25, the space E is -barrelled and hence c0- quasibarrelled. −1 (vii) ⇒ (ix): Choose (μn)n in E , and set tn := (μn+1) for each n ∈ N. The sequence (tnμn)n is uniformly bounded on the barrel [X, 1],soitisσ(E ,E)- bounded. Now, again applying Proposition 2.25, we have that (tnμn)n is equicontinuous. (ix) ⇒ (x): Let (Un)n be a sequence of neighborhoods of zero in Cd (X).Let (Kn)n be a sequence of support sets in X, and let (an)n be a sequence of positive scalars such that f ∈ C(X) : sup |f(x)|≤an := [Kn,an]⊂Un x∈Kn for each n ∈ N.Byμn we denote a positive continuous linear functional on Cc(X) whose support is the set Kn. By the assumption, there exists a sequence (tn)n of positive scalars such that (tnμn)n is equicontinuous. This implies that the series −n ∈ = n 2 tnμn has a limit μ E in the topology σ(E ,E)with support K.Iff(K) {0} for f ∈ C(X), the positive part, f +,off and the negative one, f −, vanish on K. Therefore + − μ(f ) = μ(f ) = μ(f ) = 0. Consequently, we have

+ − μn(f ) = μn(f ) = μn(f ) = 0 2.6 Pseudocompact spaces, Warner boundedness and spaces Cc(X) 55 for each n ∈ N,soKn ⊂ K. This means that [ ]⊂[ ]⊂ −1 K,1 Kn, 1 an Un −1 and hence n an Un is a neighborhood of zero in Cd (X). We proved (x). (x) ⇒ (viii): Let (Kn)n be a sequence of support sets in X. By the assumption from (x), it follows that there exists a sequence (an)n of positive scalars such that U := an[Kn, 1] n is a neighborhood of zero in Cd (X). Then there exists a compact set K in X such that [K,ε]⊂U for some ε>0. Note that Kn ⊂ K for each n ∈ N. Indeed, assume that x ∈ X \ Kn. Then there exists f ∈[X, 1] such that f(x)= 0 and f(K)={0}. Hence

(an + 1)f ∈[K,ε]⊂U ⊂[Kn,an] for each n ∈ N. This implies x/∈ Kn. Since K is compact, the closure of the set n Kn is compact in X. This proves (viii). To complete the proof, we need to show the equivalence (viii) ⇔ (xi). (viii) ⇒ (xi): Note that for a regular Borel measure μ on X there exist sequences + − + − (Kn )n, (Kn )n of compact subsets of supp μ and supp μ , respectively, such that + = + + − = − − μ (X) sup μ (Kn ), μ (X) sup μ (Kn ). n n + + − − By (viii) we deduce that the closure K of n Kn and K of n Kn are compact sets. This shows that μ+ and μ− vanish on the Borel sets that miss K+ and K−, respectively. Hence μ = μ+ − μ− vanishes on the Borel sets that miss K+ ∪ K−. Consequently, the set supp μ is a closed set of the compact set K+ ∪ K−.This proves that supp μ is compact. (xi) ⇒ (viii): Assume that (Kn)n is a sequence of nonempty support sets in X. For each n ∈ N, choose a nonnegative regular Borel measure μn on X such that μn(X) ={1} and supp μn = Kn. Set −n μ(A) := 2 μn(A). n

~~It is easy to see that μ is a nonnegative regular Borel measure on X whose support contains each Kn. Applying (xi), we deduce that the closure of n Kn is compact.~~

~~Using Theorem 2.14, we provide a version of Proposition 2.26 about (df )- spaces Cc(X).~~

Corollary 2.10 The following assertions are equivalent: (i) Cc(X) is a (df )-space. ∞ (ii) Cc(X) has a fundamental sequence of bounded sets and is -barrelled. 56 2 Elementary Facts about Baire and Baire-Type Spaces

∗ Proof (i) ⇒ (ii): By Theorem 2.14, we know that the weak dual of Cc(X) is locally ∞ complete. Then Cc(X) is -barrelled. ∞ (ii) ⇒ (i): In general, -barrelled ⇒ c0-barrelled ⇒ c0-quasibarrelled.

Recall that a topological space X satisﬁes the countable chain condition if every pairwise disjoint collection of nonempty open subsets of X is countable. We are ready to show the following interesting example.

~~Example 2.4 (KakolÐSaxonÐTodd) ˛ There exists a (df )-space Cc(X) that is not a (DF )-space.~~

Proof Recall that a point p in a topological space X is a P -point if every Gδ- set containing p is a neighborhood of p;see[181, Problems 4L]. From van Mill [297], there exists a nonÐP -point x0 in the closed subspace βN \ N such that for a sequence (Kn)n of closed subsets in βN \ N not containing x0 (and satisfying the countable chain condition) the closure of n Kn does not contain x0. Support sets satisfy the countable chain condition. Then a countable union of support sets of X := (βN \ N) \{x0} has the same closure in βN as in X and hence is compact. Applying (viii) of Theorem 2.14, we note that Cc(X) is a (df )-space. We prove that Cc(X) is not a (DF )-space. Indeed, since x0 is not a P -point of βN \ N, there N \ N exists a sequence (Un)n of open neighborhoods of x0 in β such that n Un is not a neighborhood of x0. This implies that x0 belongs to the closure of the set \ = \ X n Un n(X Un). Since each set

~~Kn := X \ Un = (βN \ N) \ Un is compact but the union n Kn has noncompact closure, by Proposition 2.26 the space Cc(X) is not a (DF )-space.~~

∞ This concrete space Cc(X) provides an example of an -barrelled space Cc(X) (by Corollary 2.10) that is not ℵ0-quasibarrelled. This space Cc(X) is not a ℵ0- barrelled space. This answers an old question posed by Buchwalter and Schmets [69].

~~2.7 Sequential conditions for locally convex Baire-type spaces~~

In this section, we collect a few results (mostly from [120] and [223]) about sequential conditions for an lcs with some Baire-type assumptions. It is well known from a classical theorem of Mahowald [280] that an lcs E is barrelled if and only if every closed linear map from E into a Banach space is continuous. It turns out that for many concrete spaces E (for example, locally complete bornological spaces E) every linear map with a sequentially closed graph from E into a Banach space is continuous. Snipes [383] characterized an lcs E (called C- sequential; see also [317]) for which every sequentially continuous linear map of E into a Banach space is continuous. De Wilde [110] showed the following. 2.7 Sequential conditions for locally convex Baire-type spaces 57

~~Proposition 2.27 (De Wilde) If E is the inductive limit of a family of metrizable Baire lcs spaces and F is a webbed space, then every sequentially closed linear map of E into F is continuous.~~

In particular, Proposition 2.27 applies if F is an (LF )-space. Recall again for convenience that an lcs E is called bornological if every absolutely convex bornivorous set in E is a neighborhood of zero if and only if every linear map from E into a Banach space that transforms bounded sets into bounded sets is continuous if and only if E is the inductive limit of a family of normed spaces. In this section, we study an lcs E having the following property (s): Every sequentially closed linear map of E into an arbitrary Fréchet space F is continuous. We shall say that an lcs E is s-barrelled [120] if it satisﬁes the property (s) above. Clearly, for a metrizable lcs the barrelledness and s-barrelledness conditions coincide. Inductive limits and Hausdorff quotients of s-barrelled spaces and are s- barrelled; in particular inductive limits of metrizable barrelled spaces and hence ultrabornological (or locally complete bornological) spaces are s-barrelled. In [223], we introduced and characterized a class of lcs called CS-barrelled,for which the condition (s) holds with Fréchet space F replaced by Banach space F . We shall say that an lcs E is CS-barrelled if every sequentially closed linear map of E into a Banach space is continuous. Clearly, the class of s-barrelled spaces is included in the class of CS-barrelled spaces. It seems to be unknown whether both classes coincide. A sequence (Un)n of absolutely convex subsets of an lcs E such that

~~Un+1 + Un+1 ⊂ Un,~~

n ∈ N,issaidtobeCS-closed if whenever n ∈ N, xm ∈ Um, m>n and x = ∞ ∈ m=n+1 xm exists in E, then x Un. The sequence (Un)n will be called absorbing if every Un is absorbing in E. Set Q = (tn)tn ≥ 0, tn = 1 . n

By a convex series of elements of a subset A ⊂ E we mean a series of the form ∈ ∈ ∈ N n tnxn, where xn A and (tn) Q, n . Following Jameson [211], [212], a subset A ⊂ E is called CS-closed if it contains the sum of every convergent convex series of its elements. Clearly, CS-closed sets are convex. Every open (or sequentially closed) convex set is CS-closed. Every convex Gδ-subset of a Fréchet space is CS-closed; see [174]. Moreover, if U ⊂ E is CS-closed, the sets 2−nU for n ∈ N form a CS-closed sequence. Let U = (Un)n be an absorbing CS-closed sequence in an lcs E. Set N(U ) = n Un, and let QU : E → E/N(U ) be the quotient map. Set EU = E/N(U ). Then the sets QU (Un) form a basis of neighborhoods of zero of a metrizable locally convex topology on E/N(U ).Let 58 2 Elementary Facts about Baire and Baire-Type Spaces

˜ EU be the completion of EU . Clearly, QU : E → EU is continuous if and only ˜ if QU : E → EU is continuous. We start with the following characterization of an lcs as s-barrelled that is due to S. Dierolf and Kakol; ˛ see [120].

Proposition 2.28 (i) For every absorbing CS-closed sequence U = (Un)n in ˜ E, the map QU as a map from E into EU is sequentially closed and Un+1 ⊂ −1 QU (QU (Un+1)) ⊂ Un, n ∈ N. Hence each Un absorbs Banach discs of E. (ii) For every Fréchet space F and every basis (Vn)n of absolutely convex closed neighborhoods of zero in F such that Vn+1 + Vn+1 ⊂ Vn, and every sequentially −1 closed linear map T : E → F , the sequence (T (Vn))n is a CS-closed sequence in E. (iii) E is s-barrelled if and only if every absorbing CS-closed sequence (Un)n of E is topological (i.e., each Un is a neighborhood of zero). ˜ Proof (i) Let (xn)n be a nullsequence in E such that QU (xn) → y in EU . Then there exists a sequence (n(k))k in N such that QU (xn(k+1) − xn(k+2)) ∈ QU (Uk+1) for k ∈ N. For any k ∈ N,set

~~yk = xn(k+1) − xn(k+2).~~

Then yk ∈ Uk+1 + N(U ) ⊂ Uk for each k ∈ N. Since (Un)n is CS-closed and ∞ ∞ xn(k+1) = ym = ym+k−1, m=k m=1 xn(k+1) ∈ Uk−1 for each k ∈ N. This implies that QU (xn(k)) → 0, so y = 0. If B is a Banach disc in E,themap ˜ QU ◦ JB :[B]→EU is closed, where [B] denotes the Banach space EB . Then QU ◦ JB is continuous by Proposition 2.27. Hence QU (B) is bounded in EU . This shows that B is absorbed by −1 QU (QU (Un+1)) ⊂ Un. ∈ N + ∈ −1 := (ii) Fix m , and for all n m 1 choose yn T (Vn) such that y ∞ k n=m+1 yn exists in E. Then ( n=m+1 T(yn))k∈N is a Cauchy sequence in F and hence converges to some z ∈ Vm. Since T is sequentially closed, z = T(y). Conse- −1 quently, y ∈ T (Vm). (iii) This is a direct consequence of the previous cases (i) and (ii).

~~By Proposition 2.28, we have the following corollary.~~

~~Corollary 2.11 An lcs E is CS-barrelled if and only if every absorbing CS-closed set in E is a neighborhood of zero. 2.7 Sequential conditions for locally convex Baire-type spaces 59~~

It turns out that the class of s-barrelled spaces is quite far from the class of barrelled spaces. We show that the space RR contains a dense Baire and bornological subspace that is not s-barrelled (see Example 2.5 below). We will need the following technical facts. Let G = (G, τ) be an lcs, let L be a sequentially closed linear subspace, and let B be a Banach disc in G.LetE = L +[B], and ﬁx a linear subspace M ⊂[B] such that L ∩ M ={0} and L + M = E. Let q : E → E/L be the quotient map, and set r = (q|M)−1.Byj : M →[B] and

~~p :[B]→[B]/([B]∩L), we denote the inclusion map and the quotient map, respectively. Then [B]∩L is closed in the Banach space [B],so[B]/([B]∩L) is a Banach space. We need the following lemma.~~

~~Lemma 2.11 The linear map~~

~~f := p ◦ j ◦ r ◦ q : (E, τ|E) →[B]/([B]∩L) is sequentially closed. If L is not closed in (E, τ|E), then f is not closed.~~

Proof First we show that the map f is sequentially closed. Indeed, let (xn)n be a null sequence in (E, τ|E), and let y ∈[B]/([B]∩L) be such that f(xn) → y in [B]/([B]∩L). Then there exist a sequence (zn)n in [B] and an element z ∈[B] such that p(zn) = f(xn), p(z) = y and zn → z in the space [B]. This implies that zn → z in the space (E, τ|E). Moreover, for each n ∈ N, one has

~~zn − r(q(xn)) ∈[B]∩L.~~

~~On the other hand, r(q(xn)) − xn ∈ L for each n ∈ N. Consequently,~~

~~zn − xn = (zn − r(q(xn))) + (r(q(xn)) − xn) ∈ L.~~

Since zn − xn → z in (E, τ|E), and L is sequentially closed, z ∈ L.Alsoz ∈[B], so y = p(z) = 0. Note also that the map f is not closed. It is enough to show that the kernel ker f of f is not closed in E; this is a simple consequence of the fact that L = ker f .

Now we are ready to present the promised example. Recall that a subspace F of an lcs E is locally dense if for every x ∈ E there exists a Banach disc B in E such that EB admits a sequence in F that converges to x in EB (or, equivalently, if for every x ∈ E there exists a sequence (xn)n in F and an increasing scalar sequence (an)n with an →∞such that an(xn − x) → 0). Note that every lcs that contains a locally dense bornological subspace is bornological; see [328, Proposition 6.2.7]. For the following example, we refer to [120].

~~Example 2.5 The space RR is s-barrelled and contains a dense subspace that is Baire bornological but is not CS-barrelled. 60 2 Elementary Facts about Baire and Baire-Type Spaces~~

Proof Set G := RR, and let R L := {(xt )t∈R ∈ R :|{t ∈ R : xt = 0}| ≤ ℵ0} be endowed with the product topology. Then L is a sequentially closed dense Baire subspace of G.LetC be the absolutely convex hull of the set {ϕ[a,b) : a,b ∈ Q,a< b}, where [a,b) is the half-closed interval and ϕ[a,b) is the characteristic function. Then the closure A of C is a Banach disc in G.Next,byB we denote the closure of C in the Banach space [A]. Then B is a Banach disc in [A] and hence in G.Let E = L +[B] with the topology of G. Now Lemma 2.11 is applied to deduce that there exists a sequentially closed discontinuous linear map of E into a Banach space [B]/([B]∩L). This shows that E is not CS-barrelled. The space E contains L as a dense subspace. Hence E is a Baire space. We prove that E is a bornological space. Indeed, by [118] the space L +[C] is a bornological subspace of G. Since [C] is locally dense in [B] with respect to the norm topology of [A], [C] is locally dense in [B]. Therefore L +[C] is locally dense in L +[B]. As every lcs that contains a locally dense bornological subspace is bornological, E is bornological. Finally, since G is the inductive limit of Banach spaces (that means G is ultrabornological), G is s-barrelled by Proposition 2.27.

Note that the space E from Example 2.5 is bornological and Baire, which is not the inductive limit of metrizable barrelled spaces. In fact, otherwise E would be CS-barrelled, which is impossible. It is well known that an lcs E is barrelled if and only if every pointwise bounded family of continuous linear maps of E into a Fréchet space is equicontinuous; see [328, Proposition 4.1.3]. One can ask if a similar result for the s-barrelledness holds if the assumption continuous linear maps is replaced by sequentially continuous linear maps. The answer is negative. In fact, if E is a bornological and barrelled space, then every pointwise bounded family of sequentially continuous linear maps of E into a Fréchet space is equicontinuous. On the other hand, the space E from Example 2.5 is bornological and barrelled and is not s-barrelled. The next example uses the following simple fact for which the proof is obvious.

~~Proposition 2.29 If F is an s-barrelled locally dense subspace of an lcs E, then E is s-barrelled.~~

~~Example 2.6 (DierolfÐKakol) ˛ There exists an bornological s-barrelled lcs that is not the inductive limit of metrizable barrelled spaces.~~

Proof Let F be an (LB)-space of Banach spaces (Fn)n that admits a bounded set not bounded in any step Fn (i.e., the space F is not regular); for such examples, we refer the reader for example to [328, Example 6.4.5]). Let E be the completion of F , and choose x ∈ E \ F for which there exists a sequence (xn)n in F that locally converges to x. Since F is s-barrelled, F is locally dense in G := F +[x] and Proposition 2.29 applies to deduce that G is s-barrelled. As F is bornological and F is locally dense in G, then G is bornological. Hence the quotient space G/[x] 2.7 Sequential conditions for locally convex Baire-type spaces 61 is s-barrelled and bornological. Let q : G → G/[x] be the quotient map. Then the restriction map q|F : F → G/[x] is continuous and bijective. We prove that G/[x] is not the inductive limit of metrizable barrelled spaces. Indeed, assume that H is a metrizable and barrrelled space, and let j : H→ G/[x] be a continuous inclusion. The map

(q|F)−1 ◦ j : H → F has the closed graph from a metrizable Baire-like space into an (LB)-space, so we apply Theorem 2.7 to deduce that this map is continuous. Since q|F is not open, G/[x] is not the inductive limit of metrizable and barrelled spaces.

~~We also refer the reader to [120, Example 2.7] to ﬁnd the proof of the following interesting fact.~~

~~Example 2.7 There exists an s-barrelled space that is not bornological.~~

Note that s-barrelled spaces enjoy some properties typical for bornological spaces. For example, we prove a MackeyÐUlam theorem for s-barrelled spaces. { : ∈ } ∈ Let Et t T be a family of lcs. Let E be the product t Et .Forx E,set = [ ] = ∈ Ex t xt . If xt 0 for every t T , then Ex is called a simple subspace of E; T see [114]. Clearly, Ex is isomorphic to R . De Wilde [114, Theorem 1] observed that a linear map of E into an lcs F is continuous if and only if its restrictions to all factor subspaces and to simple subspaces of E are continuous. Recall (see [328, Definition 6.2.21]) that a set T satisfies the MackeyÐUlam condition if no Ulam measure can be defined on it (i.e., no (0, 1)-valued measure m on the set 2T of all the subsets of T with m(T ) = 1 and m({t}) = 0 for all t ∈ T can be defined). It is known (see [328, Theorem 6.2.23]) that T satisfies the MackeyÐUlam condition if and only if RT is bornological. We shall need the following fact due to López-Pellicer [275]; see also [276], [328].

~~Lemma 2.12 For a completely regular Hausdorff space X, the space Cc(X) is bornological if and only if every weakly sequentially continuous functional on Cc(X) is continuous.~~

Proof Assume that every linear weakly sequentially continuous functional on Cc(X) is continuous. We prove that Cc(X) is bornological. It is enough to show that X is realcompact and apply Proposition 2.16. υ Fix x ∈ υX. Deﬁne ξ : Cc(X) → R by ξ(f):= f (x) for each f ∈ Cc(X).Note that ξ transforms bounded sets into bounded sets. Indeed, let (fn)n be a bounded sequence in Cc(X) such that |ξ(fn)|≥n for each n ∈ N. Since there exists y ∈ X υ = ∈ N ≥ ∈ N such that fn (x) fn(y) for each n ,wehavefn(y) n for each n , yielding 62 2 Elementary Facts about Baire and Baire-Type Spaces a contradiction. By the assumption, the map ξ is continuous. Hence there exists ε>0 and a compact set K ⊂ X such that

~~|ξ(f)|≤ε sup |f(z)| z∈K for each f ∈ C(X). Then x ∈ K; otherwise there exists g ∈ C(X) such that g|K vanishes, and gυ (x) = ξ(g) = 1, a contradiction. Hence υX = X,soX is realcompact.~~

~~Now we prove the following theorem from [120].~~

Theorem 2.15 Let {Et: t ∈ T } be a family of s-barrelled spaces. = RT (i) The product E t Et is s-barrelled if and only if is s-barrelled. (ii) RT is s-barrelled iff T satisﬁes the MackeyÐUlam condition.

Proof (i) Since RT is a minimal space (i.e., RT does not admit a weaker Hausdorff vector topology), RT is complemented in the product space E;see[328, Corol- lary 2.6.4]. Hence, since E is s-barrelled, the space RT is s-barrelled. The converse follows from De Wilde’s theorem [114, Theorem 1] mentioned above. (ii) Assume T satisﬁes the MackeyÐUlam condition. Then RT is bornological by [328, Theorem 6.2.23]. The space RT is complete. Hence RT is the inductive limit of Banach spaces. The space RT is s-barrelled by Proposition 2.27.IfT does not satisfy the MackeyÐUlam condition, RT is not bornological. Endow the set T with T the discrete topology. Then R = Cp(T ) = Cc(T ). By Lemma 2.12, there exists on RT a discontinuous sequentially continuous linear functional ξ. Hence RT is not s-barrelled.

~~Lemma 2.12 also provides the following corollary.~~

~~Corollary 2.12 Cc(X) is s-barrelled if and only if X is realcompact.~~

It is known that the completion of a bornological space need not be bornological; see [328, Notes and Remarks, p. 196]. If Z is a realcompact space whose associated kr -space Zkr is not realcompact (in [58] Blasco provided such examples), then the completion of Cc(Z) is Cc(Zkr ). By Corollary 2.12, the space Cc(Z) is s-barrelled but its completion is not. Since Cc(X) is barrelled if and only if X is a μ-space (Proposition 2.15), and there exist μ-spaces that are not realcompact, Corollary 2.12 also provides barrelled spaces Cc(X) that are not s-barrelled. Chapter 3 K-analytic and Quasi-Suslin Spaces

Abstract This chapter deals with the K-analyticity of a topological space E and N the concept of a resolution generated on E (i.e., a family of sets {Kα : α ∈ N } such = ⊂ ≤ that E α Kα and Kα Kβ if α β). Compact resolutions (i.e., resolutions N {Kα : α ∈ N } whose members are compact sets) naturally appear in many situations in topology and functional analysis. Any K-analytic space admits a compact resolution, and for many topological spaces X the existence of such a resolution is enough for X to be K-analytic. Many of the ideas in the book are related to the concept of compact resolution. We gather some results, mostly due to Valdivia, about lcs’s admitting resolutions consisting of Banach discs and their relations with the closed graph theorems. We present Hurewicz and Alexandrov’s theorems as well as the Calbrix–Hurewicz theorem, which yields that a regular analytic space X is not σ -compact if and only if X contains a closed subset homeomorphic to NN.

~~3.1 Elementary facts~~

In this section, we recall fundamental properties of K-analytic spaces and provide applications for topological vector spaces. A subset A ⊂ X of a topological space X is said to be relatively countably compact (countably compact)inX if every sequence (xk)k in A has a cluster point in X (in A), {xk : k ≥ n}=∅, A ∩ {xk : k ≥ n} = ∅ . k k The set A is called relatively sequentially compact (sequentially compact)iffor every sequence in A there exists a subsequence that converges to a point of X (of A). Clearly, A ⊂ X is countably compact if and only if each countably open cover of A admits a ﬁnite subcover. The set N of natural numbers will be endowed with the discrete topology, and the product NN is equipped with the product topology. Let S be the set of all ﬁnite sequences in N.Fors ∈ S,set

~~Σs := {σ ∈ NN : s<σ}.~~

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 63 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_3, © Springer Science+Business Media, LLC 2011 64 3 K-analytic and Quasi-Suslin Spaces

~~In the formula above, s<σmeans that the sequence σ extends s. Note that the family {Σs : s ∈ S} forms a base of open sets in NN, and Σt ⊂ Σs if and only if s~~

~~Proposition 3.1 Let X be a separable completely metrizable space such that no nonvoid open set in X is compact and the closed-open sets in X form an open base for X. Then X is homeomorphic with NN.~~

Proof Fix a complete metric on X. First note that for each >0 each nonvoid open set in X is the inﬁnite countable disjoint union of closed-open sets of diameter less than . This fact is applied to constructing disjoint open covers {Xs : s ∈ Sn} using −1 sets of diameter less than n with Xs ⊂ Xt if t

A topological space E is called K-analytic if E is the image under an upper semi-continuous (usco) compact-valued map T (called a K-analytic map) deﬁned in NN (i.e., for every α ∈ NN and every open neighborhood V of T(α)in E there exists an open neighborhood U of α such that T(U)⊂ V ). If T is deﬁned on a nonempty subset Ω ⊂ NN, the space E is called a Lindelöf Σ- space. Lindelöf Σ-spaces are often called K-countably determined or countably determined. K-analytic spaces are also known as K-Suslin spaces; see [282] and [421]. By K (E) (resp. P (E)) we denote the family of all compact subsets (resp. all parts) of a space E. We start with the following elementary fact; see [388], [345].

Proposition 3.2 A topological space E is K-analytic if and only if there exists a N map Tfrom N into K (E) satisfying two conditions: N (i) {Tα : α ∈ N }=E, where Tα = T (α). NN ∈ ∈ N (ii) If (αn)n is a sequence in converging to α and xn Tαn , for each n , the sequence (xn)n has an adherent point x ∈ Tα.

Proof Assume the conditions (i) and (ii) hold and E is not K-analytic. Hence there N N exist α ∈ N , an open neighborhood U of Tα := T(α), a sequence (αn)n in N NN converging to α in and a corresponding sequence (Uαn )n of open neighborhoods ⊂ ∈ N ∈ N of αn such that T(Uαn ) U for each n . Hence, for each n there exists ∈ \ xn T(Uαn ) U.

By (ii), the sequence (xn)n has an adherent point that belongs to Tα, and this yields a contradiction. Now assume E is K-analytic. Clearly (i) holds. Assume that (ii) fails. Then there N N exist a sequence (αn)n in N that converges to α ∈ N and a sequence xn ∈ T(αn) 3.1 Elementary facts 65 for each n ∈ N that does not have an adherent point in T(α).Fory ∈ T(α), there exist an open neighborhood V(y) of y and n(y) ∈ N such that xn ∈/ V(y) for all n ≥ n(y). Since T(α) is a compact subset of E, we can ﬁnd a ﬁnite number of points y1,y2,...,ym in T(α)such that

~~T(α)⊂ V := V(y1) ∪ V(y2) ∪ .....∪ V(ym).~~

~~The space E is K-analytic, so by deﬁnition there exists a neighborhood U of α in NN such that T(U)⊂ V . Then there exists p ∈ N, where~~

~~p ≥ max {n(y1), n(y2),...,n(ym)} such that αp ∈ U. This implies~~

~~xp ∈ T(αp) ∈ T(U)⊂ V, a contradiction.~~

~~This yields the following useful corollary [388, Theorem 2.1].~~

~~Corollary 3.1 Let ξ and β be two topologies on X such that ξ and β coincide on all ξ-compact sets. If (X, ξ) is K-analytic, (X, β) is K-analytic.~~

The following concept was introduced by Valdivia [421]. A topological space E is called quasi-Suslin if there exists a map T (called a quasi-Suslin map) from NN into P (E) satisfying: N (a) {Tα : α ∈ N }=E, where Tα = T (α). NN ∈ ∈ N (b) If (αn)n is a sequence in converging to α and xn Tαn , for each n , then the sequence (xn)n has an adherent point x ∈ Tα. Below we collect a few fundamental properties of K-analytic spaces that will be used in the subsequent parts of the book. See [346]or[421] for the proofs.

Proposition 3.3 Closed subspaces of K-analytic spaces are K-analytic. Count- able products of K-analytic spaces are K-analytic. Continuous images of K-analytic spaces are K-analytic. If a topological space E is covered by a sequence of K- analytic subspaces, E is K-analytic. Countable intersections of K-analytic subspaces of E are K-analytic.

A topological space E is called Lindelöf if one of the following equivalent conditions holds: (i) Every open cover of E has a countable subcover. (ii) For every family A of closed subsets of E with empty intersection there is a countable subfamily of A with empty intersection. (iii) Every family A of nonempty closed subsets of E that is closed under countable intersections veriﬁes A = ∅. Note the following well-known important facts; see [346]or[388], [80]. 66 3 K-analytic and Quasi-Suslin Spaces

~~Proposition 3.4 The image under a compact-valued usco map of a Lindelöf space is a Lindelöf space. Hence every K-analytic space is Lindelöf.~~

Proof Let F be a Lindelöf space, and let T be a compact-valued usco map from F onto a topological space E.Let(Vi)i∈I be an open cover of E.Forx ∈ F , there is a ⊂ ﬁnite subset I(x)of I such that T(x) i∈I(x)Vi. Since T is usco, there exists an open neighborhood U(x) of x such that

~~T(U(x))⊂ Vi. i∈I(x)~~

~~As F is Lindelöf, there exist a sequence(xn)n in F and a sequence (U(xn))n of ⊂ open neighborhoods of xn such that F n U(xn). Hence~~

~~E = T(F)⊂ T(U(xn)) ⊂ Vi, n n i∈I(xn) and E is Lindelöf.~~

W = : ∈ N ≤ ≤ ∈ N Following De Wilde, a web Ca1a2...ak ai , 1 i k,k inasetE is called ordered (see [421]) if for integers k,a1,a2,...,ak and b1,b2,...,bk such ≤ ≤ ≤ ⊂ that aj bj ,1 j k,wehaveCa1a2...ak Cb1b2...bk .

Theorem 3.1 (Cascales) The following conditions are equivalent for a topological space E: (i) E is a quasi-Suslin space. (ii) There is a quasi-Suslin ordered map A : NN −→ P (E) (i.e., given α, β ∈ NN such that α ≤ β, then Aα ⊂ Aβ ). W = : ∈ N ≤ ≤ ∈ N (iii) There is an ordered web Ca1a2...ak ai , 1 i k,k such = ∈ NN ∈ ∈ N that, for every α (ak) , if xk Ca1a2...ak , k , the sequence (xk)k has an adherent point in E that belongs to k Ca1a2...ak .

~~Proof (i) ⇒ (ii): Let T : NN → P(E) be a quasi-Suslin map. For α ∈ NN,set~~

~~N Aα := {Tβ : β ≤ α, β ∈ N }.~~

Clearly, Aα ⊂ Aβ if α ≤ β. In order to show that A : α → Aα is quasi-Suslin, we need only to check condition (b) for the map A : α → Aα. j N N Let (αj ), with αj := (an), be a sequence in N that converges to α = (an) ∈ N , ∈ ∈ N and assume that xj Aαj for each j . By the deﬁnition of the sets Aα, it follows ∈ N ≤ ∈ N that for each j there exists βj αj such that xj Tβj . Since is a discrete jk space, there exists an increasing sequence (jk)k of elements of N such that an = an ∈ N NN for all k,n . The subsequence (βjk )k of (βj )j is relatively compact in ,so (β ) contains a subsequence (β ) that converges to β ∈ NN and β ≤ α.Asthe jk k jkl l map T is a quasi-Suslin map, condition (b) is satisﬁed, so the subsequence (x ) jkl l 3.1 Elementary facts 67 of (xj )j has an adherent point x ∈ Tβ . This implies also that the sequence (xj )j has N an adherent point belonging to Aα. The condition (b) for the map A : N → P(E) is proved. N (ii) ⇒ (iii): For α = (nk) ∈ N ,set

:= { : = ∈ NN = ≤ ≤ } Cn1n2.....nk Aα α (ak) ,aj nj , 1 j k . { } { } Then Cn1n2.....nk k is an ordered web in E. We claim that Cn1n2.....nk k satisﬁes the = ∈ NN ∈ conditions claimed in (iii). Indeed, take α (ak) and select xk Cn1n2.....nk ∈ N = k ∈ NN for each k . By the deﬁnition of Cn1n2.....nk , one gets αk (an) such that k = ∈ = aj aj , and xk Aαk for j 1, 2,...k. Then the sequence (αk)k converges to α. Consequently, the sequence (xk)k has an adherent point in E that belongs to ⊂ Aα Cn1n2.....nk . k

~~N (iii) ⇒ (i): Deﬁne the map T : N → P(E) by T(α):= Tα, where := Tα Cn1n2.....nk k~~

N for α = (ak) ∈ N . This is a quasi-Suslin map. Indeed, let (αk) be a sequence in NN = k = ∈ NN ∈ ∈ with αk (an) that converges to α (an) , and let xj Tαj for all j N . There exists a subsequence (xjn )n of (xj )j with an adherent point x in E that belongs to Tα by the assumption (iii). Since x is also an adherent point of (xj )j ,the conclusion holds.

~~A similar argument applies to prove the following result of Talagrand [388]; see also [345], [80].~~

Theorem 3.2 (Talagrand, Rogers) The following conditions are equivalent for a topological space E: (i) E is K-analytic. N (ii) There is a K-analytic ordered map A : N −→ K (E) . W = : ∈ N ≤ ≤ ∈ N (iii) There is an ordered web Ca1a2...ak ai , 1 i k,k in E = ∈ NN ∈ such that for every α (ak) the set k Ca1a2...ak is compact, and if xk ∈ N Ca1a2...ak , for k , the sequence (xk)k has an adherent point in E belonging to k Ca1a2...ak .

We complete this section with some additional information about Lindelöf Σ- spaces. We recall another approach to deﬁning the class of Lindelöf Σ-spaces due to Nagami [308]. A topological space X is a Lindelöf Σ-space if X is a Σ-space (in the sense of [308]) and Lindelöf; see [21], [256]. Proposition 3.5 below gath- ers several known characterizations of Lindelöf Σ-spaces; see also [256], [346], [25], [318]. We refer the reader to a recent paper by Tkachuk [402] about Lindelöf Σ-spaces. 68 3 K-analytic and Quasi-Suslin Spaces

A family N in a topological space X is a network of X if for every x ∈ X and every open neighborhood U of x there exists V ∈ N with x ∈ V ⊂ U. Recall that a family N of subsets of X is called a network with respect to a cover C of X if for every set C ∈ C and every open neighborhood U of C there exists N ∈ N such that C ⊂ N ⊂ U.

Proposition 3.5 Let X be a completely regular Hausdorff space. The following assertions are equivalent: (i) X is a Lindelöf Σ-space. (ii) There exist a compact cover C and a countable network N with respect to C . (iii) There exist a metrizable and separable space M and a compact-valued usco map T such that T(M)= X. (iv) There exists a compact cover C and a countable network G with respect to C consisting of closed sets. (v) There is a compact-valued usco map from a subset of NN. (vi) There exists a countable family F of compact subsets of βX such that F separates X (i.e., for any x ∈ X and y ∈ βX \ X there exists F ∈ F such that x ∈ F and y/∈ F ). (vii) X is a continuous image of a separable zero-dimensional metrizable space. (viii) There exist a metrizable separable space M, a space L and maps g : L → M and f : L → X such that g is perfect and f is surjective and continuous. (ix) There are a metrizable and separable space M, a compact space K, aclosed subspace F of M × K and a continuous map f : F → X such that f(F)= X.

Proof We prove only the equivalences (ii) ⇔ (iii) and (ii) ⇔ (iv) ⇔ (v) ⇔ (vi). For the remaining part of the proof, we refer to [256], [346]. (ii) ⇒(iii): Let N ={Ni}i . For each C ∈ C , determine a sequence (cn)n such = := { : ∈ C } that C n Ncn .LetM (cn)n C . Then the map deﬁned in M by := T ((cn)n) Ncn n is compact-valued, usco and T(M)= X. (iii) ⇒ (ii): Let T : M → X be a compact-valued usco map onto X.LetB be a countable basis for M. Then {T(B): B ∈ B} is a countable network with respect to the compact cover {T(t): t ∈ M} of X. (v) ⇒ (ii): Let φ : Σ → X be a usco compact-valued map from a subset Σ of NN onto X.LetB be a countable base in Σ that is closed under the ﬁnite unions. Set N := {φ(B): B ∈ B}. Then C := {φ(x): x ∈ Σ} is a compact cover of X, and N is a countable network with respect to C . (ii) ⇒ (iv): Let C and N be as in (ii). Set

~~G := {N : N ∈ N }.~~

It is enough to show that G is a network with respect to C .FixC ∈ C , and let U be an open neighborhood of C. By the regularity of X,forx ∈ C, there exists an 3.1 Elementary facts 69 open neighborhood Vx of x such that Vx ⊂ U. Since C is compact, there exists ﬁnite D ⊂ C such that

~~C ⊂ V := Vx. x∈D Take N ∈ N with C ⊂ N ⊂ V . Then~~

~~C ⊂ N ⊂ V ⊂ U.~~

~~This provides (iv). (iv) ⇒ (vi): Let C and G be as in (iv). Then~~

F := {N : N ∈ G } is countable, where the closure is taken in βX.Letx ∈ X and y ∈ βX \ X. There exists C ∈ C with x ∈ C.LetW be an open neighborhood of C in βX with y/∈ W . Then there exists N ∈ G such that

~~C ⊂ N ⊂ W ∩ X.~~

Consequently, x ∈ C ⊂ N ⊂ W ∩ X ⊂ W ⊂ βX \{y}. Hence F := N ∈ F and x ∈ F and y/∈ F . (vi) ⇒ (v): Let F = (Kn)n be as in (vi). Set := = ∈ NN : ⊂ Σ α (an) Kan X . n ∈ := : → For each α Σ,thesetφ(α) n Kan is compact and φ Σ X is a compact- N valued map. Let x ∈ X. Choose α = (an) ∈ N such that

~~{n ∈ N : x ∈ Kn}={an : n ∈ N}. \ := ⊂ F separates X from βX X,soF n Kan X. Hence~~

~~α := (an) ∈ Σ, x ∈ F = φ(α), φ(Σ)= X.~~

We need to show that φ is a usco map. Fix arbitrary α ∈ Σ, and let U be an open neighborhood of the compact set φ(α).LetV be an open set in βX with V ∩X = U. ∈ N ⊂ There exists m such that n≤m Kan V . Then

~~G := {β = (bn) ∈ Σ : bn = an,n≤ m} is open in Σ and φ(G)⊂ U.~~

~~The ﬁrst consequence of the previous result (part (v)) is the following. 70 3 K-analytic and Quasi-Suslin Spaces~~

~~Corollary 3.2 Any σ -compact space X (i.e., X covered by a sequence of compact sets or, more generally, every K-analytic space) is a Lindelöf Σ-space.~~

~~Corollary 3.3 Any topological space X with a countable network is a Lindelöf Σ-space.~~

~~Proof A countable network is a network with respect to the compact cover {{x}: x ∈ X}. Now apply Proposition 3.5(ii).~~

~~We note also the following interesting consequence of Theorem 3.5;see[402, Theorem 2.14] or [27, Proposition IV.6.15].~~

~~Corollary 3.4 If for a Lindelöf Σ-space X every compact subset of X is ﬁnite, X is countable.~~

Proof By Theorem 3.5 (ii) there exist a compact cover C and a countable network N C N := { ∈ N : with respect to that is closed under finite intersections. Set 0 N N is finite}. Assume X is uncountable. Since M := N is countable, there N∈N0 exists x ∈ X \ M.FixC ∈ C with x ∈ C. There exists a countable decreasing sequence (Pn)n in N such that, for each open neighborhood U of C, there exists n ∈ N such that C ⊂ Pn ⊂ U.Asx/∈ M,everyPn is infinite. Since C is finite (by the assumption), we can find an injective sequence (xn)n with

~~xn ∈ Pn \ C for each n ∈ N. Set D := {xn : n ∈ N}. Since D \ U is ﬁnite for each open neighborhood of C, C ∪ D is an inﬁnite compact subset of X, a contradiction. We proved that X = M.~~

~~We note also the following useful proposition.~~

~~Proposition 3.6 Let X be a K-analytic space endowed with a web~~

N := { : ∈ N ∈ N} Cn1,n2,...,nk nk ,k := = ∈ NN C := { : as in Theorem 3.2. Set Uα k Cn1,n2,...,nk for α (nk) . Then Uα α ∈ NN} is a compact cover of X, and N is a countable network with respect to C .

⊂ ⊂ Proof Let U be an open set in X, and assume that Uα U. Since Uα Cn1,n2,...,nk ∈ N ∈ N ⊂ for each k , it is enough to show that there exists k such that Cn1,n2,...,nk ∈ N ∈ ∈ U. Assume that for each k there exists xk Cn1,n2,...,nk such that xk / U. Then ∈ = (xk)k has an adherent point x k Cn1,n2,...,nk Uα. This provides a contradiction since Uα ⊂ U.

~~We shall need the following result due to Talagrand [388, Theorem 2.4]. 3.2 Resolutions and K-analyticity 71~~

Lemma 3.1 (Talagrand) Let (X, τ) be a regular space that admits a ﬁner topology ξ such that (X, ξ) is a Lindelöf Σ-space. Then dens(X, ξ) ≤ ω(X,τ), where dens(X) and ω(X) denote the density and weight of X, respectively.

Proof Let {ωi : i ∈ I} be a basis of open sets in τ of cardinality ℵ. Since (X, ξ) is a Lindelöf Σ-space, there exists a nonempty subset Σ of NN and a compact-valued usco map T from Σ covering (X, ξ). Set T(α):= Kα, and as usual

:= { : = ∈ = ≤ ≤ } Cn1,...,nk Kβ β (mj ) Σ,mj nj , 1 j k , ∈ ∩ ∈ ∈ N for α Σ. For every non-empty set ωi Cn1,...nk , where i I , k,nk , and (nk)k ∈ Σ, choose a single point, and let H be the set of such points. Then |H |≤ℵ. Observe that H is dense in (X, ξ). Indeed, let U be an open set in ξ, and ﬁx x ∈ U. There exist α = (nk) ∈ Σ such that x ∈ Kα, and i ∈ I such that

x ∈ ωi ∩ Kα ⊂ ωi ∩ Kα ⊂ U, where the closure is taken in τ . Moreover, there exists k ∈ N such that ∈ ∩ ⊂ x ωi Cn1,...,nk U. Indeed, otherwise there exists a sequence ∈ ∩ \ xk ωi Cn1,...,nk U for k ∈ N. Since T is usco, (xk)k has an accumulation point

~~y ∈ ωi ∩ Kα \ U, a contradiction. This proves that H ∩ U is nonempty, so H is dense in (X, ξ).~~

Note also that the class of Lindelöf Σ-spaces is the minimal class that contains all second-countable spaces, all compact spaces and is closed with respect to the ﬁnite products, closed subspaces and continuous images; see [27].

~~3.2 Resolutions and K-analyticity~~

Theorem 3.2 shows also that every K-analytic space E admits a covering {Aα : N α ∈ N } of compact subsets of E such that Aα ⊂ Aβ for α ≤ β. Such an ordered covering will be called a compact resolution of E. If the sets Aα are countably compact or relatively countably compact, the reso- N lution {Aα : α ∈ N } will be called countably compact or relatively countably compact, respectively. Let us start with the following simple observation concerning resolutions used in the subsequent parts of the book. The argument in the proof is essentially due to Talagrand; see [388, Proposition 6.13] and also Corollary 13.1 below. 72 3 K-analytic and Quasi-Suslin Spaces

~~N Proposition 3.7 Let {Aα : α ∈ N } be a resolution on an uncountable set A. Then, N for some γ ∈ N , the set Aγ is inﬁnite.~~

~~Proof We can choose a sequence (kn)n of positive integers such that the set~~

~~N Bn = {Aα : α = (ak) ∈ N ,ai = ki 1 ≤ i ≤ n} is uncountable for every n ∈ N. For each n ∈ N, choose~~

~~xn ∈ Bn \{xi : 1 ≤ i~~

~~Proposition 3.7 can also be deduced from Corollary 3.5; compare also Corollary 3.4. Proposition 3.7 yields the following.~~

~~Proposition 3.8 Let E be an uncountable-dimensional vector space, and let E∗ be the algebraic dual of E. Then neither (E, σ (E, E∗)) nor (E∗,σ(E∗,E)) is quasi- Suslin.~~

~~Proof Let {xk : k ∈ ω1} be a linearly independent set in E, where ω1 is the set of all countable ordinals. For each k ∈ ω1, where k ≥ ω0, choose a bijection~~

~~Tk :{l ∈ ω1 : l ≤ k}→N.~~

~~Since {xk : k ∈ ω1} is linearly independent, for each l ∈ ω1 there exists a linear ∗ functional ul ∈ E such that~~

~~ul(xk) = Tk(l), if k ≥ max (ω0,l).~~

~~If S is a countable subset of ω1, and k = sup S,theset~~

{ul(xk) = Tk(l) : l ∈ S} consists of distinct points in N and is unbounded in N. This implies that H := ∗ {ul : l ∈ ω1} consists of ℵ1-distinct points, and no infinite subset of H is σ(E ,E)- bounded. Therefore no infinite subset of H is contained in a σ(E∗,E)-countable compact set. Assume (E∗,σ(E∗,E)) is quasi-Suslin. Then (E∗,σ(E∗,E)) has a resolution N N {Kα : α ∈ N } of countably compact sets. By Proposition 3.7, there exists α ∈ N such that H ∩ Kα is infinite, a contradiction. ∗ For the other case, for each infinite k ∈ ω1 there exists a linear functional vk ∈ E such that

~~vk(xl) = Tk(l), if k ≥ max (ω0,l). 3.2 Resolutions and K-analyticity 73~~

~~Using the argument above, we note that (E, σ (E, E∗)) is not a quasi-Suslin space as well.~~

One cannot prove or disprove in the usual set theory a similar statement that a set with a resolution consisting of countable sets is countable (see Proposition 3.9 below). This also refers to the question of whether a Banach space that has a resolution of separable sets is separable. Indeed, note that under CH the nonseparable Banach space c0[0,ω1) has a resolution consisting of closed separable subspaces of the form c0[0,μ)with μ<ω1; see also [131]. N ∗ For α = (nk) and β = (mk) ∈ N , the relation α ≤ β means that there exists m ∈ N such that nk ≤ mk for all k ≥ m. We note the following fact from [399, Theorem 3.6].

N Proposition 3.9 Under CH the space ω1 has a compact resolution {Kα : α ∈ N } such that every compact set in ω1 is contained in some Kα, and under MA+¬CH the space ω1 does not admit a compact resolution.

Proof As each countable subset in (NN ≤∗) has an upper bound, by the transfinite induction we determine an injective map Ψ from ([0,ω1), ≤) into a well-ordered subset M of (NN ≤∗) that preserves the order. N If we assume CH, there exists a bijection ϕ between [0,ω1) and N . Then, we may assume that ϕ(t) ≤ Ψ(t)for each t ∈[0,ω1), which implies that M is a cofinal N ∗ N subset in (N ≤ ).Forα ∈ N , we denote by tα the first element in [0,ω1) such ∗ that α ≤ Ψ(tα) and let Kα := [0,tα]. Clearly, KΨ(δ) =[0,δ]. N The fact that {Kα : α ∈ N } is a compact resolution swallowing compact sets follows from the fact that for a compact subset K in [0,ω1) there exists δ ⊂[0,ω1) such that K ⊂[0,δ]=KΨ(δ). Since ω1 is noncompact but countably compact, it is not Lindelöf and hence is not K-analytic. Now observe that under MA +¬CH the space [0,ω1) does not have a compact N resolution. Indeed, assume that [0,ω1) admits a compact resolution {Kα : α ∈ N }. N If α<ω1, there exists tα ∈ N such that α ∈ Kt . By the assumption MA +¬CH, N ∗ α one gets t ∈ N with tα ≤ t for all α<ω1. Set N Wt := {q ∈ N :∃m ∈ N : q(n) = t(n),n≥ m}.

Then Wt is countable. Also, for each α<ω1 there exists q ∈ Wt such that tα ≤ q. This proves that {Kq : q ∈ Wt } covers ω1, and hence some set K must be uncountable. Since all compact sets in ω1 are countable, we have reached a contradiction.

It is easy to see that any separable metric and complete space E has a compact N resolution that swallows compact sets. Indeed, for α = (nk) ∈ N , and for a countable and dense sequence (xn)n in E,set

~~∞ nk −1 Kα := B(xj ,k ), k=1 j=1 74 3 K-analytic and Quasi-Suslin Spaces~~

−1 −1 where B(xj ,k ) is the closed ball in E with the center at point xj and radius k , N j,k ∈ N. Then {Kα : α ∈ N } is a compact resolution, and for every compact set K N in E there exists α ∈ N such that K ⊂ Kα. The following observation is due to Tkachuk [399].

Proposition 3.10 (i) Any K-analytic space has a compact resolution. (ii) Any continuous image of a compact resolution is a compact resolution. (iii) A closed subset of a space with a compact resolution has a compact resolution. (iv) A countable union of subspaces of a topological space with a compact resolution has a compact resolution. (v) A countable product of spaces with a compact resolution has a compact resolution. (vi) A countable intersection of subspaces of a topological space with a compact resolution has a compact resolution.

Proof (i) follows from Theorem 3.2. Conditions (ii) and (iii) are clear. (iv) Let (En)n be a sequence of subspaces of a space E covering E, and assume { n : ∈ NN} = ∈ NN that each En has a compact resolution Kα α in En.Forα (ak) ,the := n { : ∈ NN} set Aα 1≤n≤a Kα is compact, and Aα α is a compact resolution on E. = 1 (v) Let E n En be the countable product of topological spaces En endowed ∈ N { n : ∈ NN} with the product topology, and for each n let Aα α be a compact N resolution in En.Let(Dn)n be a division of the space into disjoint inﬁnite sets N = : N → such that n Dn.Ifξn Dn is a bijection, the map

∗ N : NDn → N ξn deﬁned by ∗ := ◦ ξn (f ) f ξn ∈ N ∗ ≤ ∗ ≤ for each n is a bijection and such that ξn (f ) ξn (g) if and only if f(n) g(n) ∈ NDn | := = ∈ NN for f,g .Ifα n (ak)k∈Dn , where α (ak) ,theset n Kα := A ∗ | ξn (α n) n

~~N is compact, and {Kα :∈ N } is a compact resolution in E. (vi) The intersection is homeomorphic to the diagonal of the product of the subspaces.~~

Since every Polish space has a compact resolution swallowing compact sets, Proposition 3.10 yields that every Cech-completeˇ Lindelöf space E admits a compact resolution swallowing compact sets. Indeed, E is homeomorphic to a closed subset of the product M × K for some Polish space M and a compact Hausdorff space K. 3.2 Resolutions and K-analyticity 75

Note that there exist (even locally convex) spaces with a compact resolution that are not K-analytic; see [388], [80], [157], [159]. We provide such examples in the sections below. A subset A of a space E is called full if A contains the adherent points of the sequences in A. Note the following simple fact from [157];seealso[80].

~~N Proposition 3.11 Assume that E admits a resolution {Aα : α ∈ N } consisting of countably compact full sets. Then E is quasi-Suslin.~~

~~N Proof For α = (an)n ∈ N ,set~~

= : ∈ NN | = | Ca1a2...ak Aβ β ,β k α k and Bα = ∈N Ca a ...a , where as usual we denote α|k := (a1,a2,...,ak) for k 1 2 k N each k ∈ N. We prove that the map T , deﬁned by T(α)= Bα,forα ∈ N ,is N N a quasi-Suslin map. Indeed, if α = (ak) ∈ N , and βk ∈ N , βk|k = α|k and ∈ ∈ N ∈ NN xk Bβk for each k , then, by the order condition, there exists γk such { : ≥ } ⊂ | = | ∈ N that xn n k Aγk and γk k α k for each k . The sequence (xn)n has an ∈ ⊂ adherent point x Aγ1 Ca1 .Asx is also an adherent point of (xn)n≥k, and (xn)n≥k ∈ ⊂ ≥ is contained in the full set Aγk , we have thatx Aγk Ca1a2...ak for k 1. Then the ∈ = sequence (xn)n has an adherent point x k Ca1a2...ak Bα, implying that T is a quasi-Suslin map.

~~We shall also need the following well-known property for Lindelöf spaces; see [146].~~

~~Proposition 3.12 Every Lindelöf space E is realcompact.~~

−1 Proof If E is not realcompact, there exists y ∈ βE\E such that Zf := f (0)∩E is a closed nonvoid subset of E for each f ∈ F := {f∈ C(βE) : f(y)= 0,f(βE)⊂ [ ]} −n ∈ F 0, 1 (see Proposition 2.20). By the continuity of n 2 fn,forfn , we know that the family {Zf : f ∈ F } satisﬁes the countably intersection property. Since {Zf : f ∈ F }=∅, we deduce that E is not Lindelöf.

~~We provide another useful characterization for a space to be K-analytic; see [181].~~

Proposition 3.13 For a regular topological space E, the following assertions are equivalent: (i) E is K-analytic. (ii) E is quasi-Suslin and Lindelöf. (iii) E is Lindelöf and admits a compact resolution. (iv) E is Dieudonné complete (i.e., E is a closed subspace of a product of metrizable spaces) and admits a compact resolution. (v) Every relatively countably compact set in E is relatively compact, and E admits a compact resolution. 76 3 K-analytic and Quasi-Suslin Spaces

Proof (i) ⇒ (ii): Clearly, any K-analytic space is quasi-Suslin. By Proposition 3.4, every K-analytic space is Lindelöf. (ii) ⇒ (i): If T is a quasi-Suslin map, each set T(α)is compact. (i) ⇒ (iii) is clear. (iii) ⇒ (iv): Every Lindelöf space is realcompact and thus homeomorphic to a closed subspace of a product of the real lines. (iv) ⇒ (v): Every relatively countably compact set in a Dieudonné space is relatively compact. ⇒ { N (v) (i): Let Kα : α ∈ N } be a compact resolution in E, and set Bα := = ∈ NN k Cn1,n2,...,nk for α (nk) . Since Bα is countably compact, Bα is compact. Now it is clear that the map K(α) := Bα is usco. Hence E is K-analytic.

~~Corollary 3.5 follows from [399].~~

~~Corollary 3.5 (Tkachuk) Every topological space E having a compact resolution has a countable extent (i.e., every closed discrete subset of X is countable).~~

Proof Let D be a closed and discrete subset of E. Then D with the induced topology has a compact resolution. Proposition 3.13(v) can be applied to deduce that D is K- analytic and hence Lindelöf, so D is countable.

The following concept is due to Fremlin (see [165]). A Hausdorff topological space E is called angelic if every relatively countably compact set K in E is relatively compact and for every x ∈ K there exists a sequence in K converging to x. In angelic spaces, the (relatively) countably compact, (relatively) compact and (relatively) sequentially compact sets are the same; see [165, Theorem 3.3], and also [185], where a proper subclass of angelic spaces has been studied. Classical examples of angelic spaces include spaces Cp(X) with compact X (see, for example, [190], [240]), Banach spaces E with the weak topology σ(E,E) and spaces of ﬁrst Baire class functions on any Polish space P with the topology of the pointwise topology on P (see [71]). Below we prove a general result due to Orihuela, providing a much larger class of angelic spaces Cp(X) and applications. If a topological space E admits a compact resolution, E is quasi-Suslin by Propo- sition 3.11. Nevertheless, E need not be K-analytic since the countably compact sets Bα, obtained in the proof of Proposition 3.11, need not be compact. For angelic spaces, countably compact sets are compact, so we note the following facts; see [80].

~~Corollary 3.6 For an angelic space E, the following statements are equivalent: (i) E admits a compact resolution. (ii) E is a quasi-Suslin space. (iii) E is a K-analytic space.~~

~~Proof (i) ⇒ (ii): This follows from Proposition 3.11. (ii) ⇒ (iii) since every countably compact set is compact in any angelic space. 3.2 Resolutions and K-analyticity 77~~

~~(iii) ⇒ (i) since every K-analytic space admits a compact resolution.~~

~~From Proposition 3.11, we deduce also the following corollary.~~

W ={ ∈ N ≤ ≤ ∈ N} Corollary 3.7 Let Dn1n2...nk ,ni , 1 i k, k be an ordered web = NN = in a topological space E. If for every α (ak)k in the set Aα k Da1a2...ak is countably compact (compact), E is quasi-Suslin (K-analytic).

~~Proof From Proposition 3.11, it follows that E is quasi-Suslin. As usual, deﬁne~~

:= : ∈ NN | = | Ca1a2...ak Aβ β ,β k α k = ∈ NN := for each α (ak)k , and let Bα k Ca1a2...ak . Then the map T deﬁned := ∈ NN ⊂ by T(α) Bα for α is a quasi-Suslin map. Obviously, Aα Ca1a2...ak (see ∈ NN ⊂ Proposition 3.11 and the proof) and, if β , then Aβ Dβ1β2...βk . Therefore, if | = | ⊂ β k α k, then Aβ Da1a2...ak . This implies ⊂ Ca1a2...ak Da1a2...ak .

~~Since ⊂ ⊂ Aα Ca1a2...ak Da1a2...ak , for each k ∈ N, we have ⊂ ⊂ = Aα Bα Da1a2...ak Aα. k~~

~~Since Aα = Bα,themapT is a K-analytic map in E if each Aα is compact. Conse- quently, E is K-analytic.~~

For a locally convex space E,bytheMackey topology we mean the ﬁnest locally convex topology μ(E, E) on E having the same continuous linear functionals as the original topology of E. By the Mackey–Arens theorem (see [213, Theorem 8.5.5]), the topology μ(E, E) is the topology of the uniform convergence on absolutely convex σ(E,E)-compact subsets of E. An lcs is called quasicomplete if every closed bounded set in E is complete [213]. For the following application of Proposition 3.11, we refer to [80].

Corollary 3.8 Let E be an lcs that is quasicomplete for the Mackey topology μ(E, E). Then the following statements are equivalent: (i) E admits a compact resolution. (ii) E is a quasi-Suslin space. (iii) E is a K-analytic space.

~~Proof From Proposition 3.11, it follows that (i) ⇒ (ii). 78 3 K-analytic and Quasi-Suslin Spaces~~

~~(ii) ⇒ (iii): Assume E is quasi-Suslin, and let T1 be an ordered quasi-Suslin map N deﬁned by T1(α) := Bα, where {Bα : α ∈ N } is a countably compact resolution. N For α = (nk) ∈ N ,set~~

= : ∈ NN | = Dn1...nk Bβ β ,β k (n1 ...nk) := := and Aα k∈N Dn1n2...nk . It is clear that the map T deﬁned by T(α) Aα is a quasi-Suslin map. The regularity and the accumulation property imply = Aα Dn1n2...nk . k∈N

~~By the assumption on E, the closure Aα of the countable compact set Aα is compact (see [246, Theorem 24.2.1]), and then Corollary 3.7 applies. (iii) ⇒ (i) is obvious.~~

~~We note another application of Corollary 3.7;see[80].~~

Corollary 3.9 Let E be a semireﬂexive lcs. The following assertions are equivalent: N (i) E admits a bounded resolution {Aα : α ∈ N } (i.e., the sets Aα are bounded in the locally convex space E). (ii) (E, σ (E, E)) is K-analytic. (iii) (E, σ (E, E)) is quasi-Suslin.

N N Proof (i) ⇒ (ii): Let {Aα : α ∈ N } be a bounded resolution. Let α = (an) ∈ N . Then, for every closed absolutely convex neighborhood of zero U in E, there exists k ∈ N such that ⊂ k Ca1a2.....ak 2 U. Indeed, otherwise there exists a neighborhood of zero U in E such that for every ∈ N ∈ −k ∈ ∈ k there exists xk Ca1a2.....ak such that 2 xk / U. Since xk Ca1a2.....ak for ∈ N = k ∈ NN ∈ k = = every k , there exists βk (bn)n such that xk Aβk and bj aj for j 1, 2,...,k. Set = { k : ∈ N} dn max bn k ∈ N = ≥ ∈ N ⊂ ∈ for n . Set γ (dn). Since γ βk for every k , then Aβk Aγ ,soxk Aγ ∈ N −k → for all k .AsAγ is bounded, 2 xk 0inE, a contradiction. This implies := that the set W k Ca1a2.....ak is bounded and closed in E, where the closure is taken in σ(E,E). Since E is semireﬂexive, the set W is compact in σ(E,E).Now Corollary 3.7 applies to get that (E, σ (E, E)) is K-analytic. (ii) ⇒ (iii) is clear. (iii) ⇒ (i): Since every quasi-Suslin space admits a resolution of countably compact sets and every such set is bounded, the conclusion follows.

~~The following proposition for Banach spaces E is due to Talagrand [388, The- orem 3.6]. The same statement for a Fréchet space E was proved by Canela [79, 3.2 Resolutions and K-analyticity 79~~

Proposition 4]. The argument for the proof of Proposition 3.14 is adopted from [80, Proposition 7]. Proposition 3.14 fails in general: RR is not K-analytic and RR,asa separable space, contains a dense countable-dimensional K-analytic subspace.

~~Proposition 3.14 Let E be a Fréchet space, and let F be a dense subspace of E such that (F, σ (F, F )) is K-analytic. Then (E, σ (E, E)) is K-analytic.~~

Proof By Corollary 3.8, it is enough to show that (E, σ (E, E)) has a compact N resolution. Assume {Aα : α ∈ N } is a compact resolution in (F, σ (F, F )).Let (Un)n be a decreasing basis of closed absolutely convex neighborhoods of zero in = ∈ NN := { : ∈ NN} E.Forα (nk) ,setBα k nkUk. Then Bα α is a resolution consisting of bounded complete sets, and for each bounded set B ⊂ E there exists N α ∈ N such that B ⊂ Bα. Set

~~= : ∈ NN | = Cn1...nk Bβ β ,β k (n1 ...nk) .~~

N For α = (nk) ∈ N ,set := + ∩ + −1 ∩···∩ Dα (Aα Cn1 ) A(n2,n3,... ) 2 Cn1n2 + −1 ∩ A(nk,nk+1,... ) k Cn1n2....nk ..., where the closure is taken in (E ,σ(E ,E )). We show that every Dα is contained N N in E being σ(E,E )-compact, and that E = {Dα : α ∈ N }.Fixα = (nk) ∈ N . Let U be a σ(E,E)-closed neighborhood of zero in the strong topology β(E,E) ∈ N −1 ⊂ of E . Note that there exists k such that k Cn1n2....nk U. Moreover, Dα ⊂ {E + U} for any neighborhood of zero U in (E,β(E,E)). Since E is complete, we have Dα ⊂ {E + U}=E, if U runs over a basis of neighborhoods of zero in (E,β(E,E)). This shows that = + ∩ ∩ + −1 ∩ ∩···∩ Dα Aα Cn1 E (A(n2,n3,... ) 2 Cn1n2 E) + −1 ∩ ∩ A(nk,nk+1,... ) k Cn1n2.....nk E ....

Consequently, every set Dα is bounded in E. Since every Dα is an intersection of σ(E ,E )-closed subsets of E ,wehavethatDα is σ(E ,E )-closed, so it is a σ(E,E)-compact subset of E. Now ﬁx any x ∈ E. Then there exists a sequence N (xn)n in F such that n(xn − x) → 0. Choose α = (ak) ∈ N such that {n(xn − x) : ∈ N}⊂ ⊂ n Bα. Since Bα k Ca1....ak ,wehave − ∈ −1 x xk k Ca1....ak 80 3 K-analytic and Quasi-Suslin Spaces

∈ N = ∈ NN ∈ ∈ N for all k .Letβ (bj ) be such that xk A(bk,bk+1,... ) for each k . Let cj := max{aj ,bj } for all j ∈ N. Then x ∈ Dα if α = (cj ). We proved that N {Dα : α ∈ N } is a compact resolution on (E, σ (E, E )).

~~We conclude this section with the following result from [88] that generates special usco maps.~~

Theorem 3.3 (Cascales–Kakol–Saxon) ˛ Let X be a ﬁrst-countable topological space. Let Y be a regular topological space for which the relatively countably com- Y pact subsets are relatively compact. Let φ : X → 2 be a set-valued map such that n∈N φ(xn) is relatively compact for each convergent sequence (xn)n in X. If for each x ∈ X we set ψ(x):= {φ(V): V neighborhood of x in X}, (3.1) the map ψ : X → 2Y is usco and satisﬁes φ(x)⊂ ψ(x) for every x ∈ X.

~~Proof For x ∈ X,set~~

~~C(x) := {y ∈ Y :∃xn → x, ∀n ∃ yn ∈ φ(xn), y cluster of (yn)n}. (3.2)~~

~~x Fix (Vn )n, a decreasing basis of open neighborhoods of x.~~

Claim 3.1 C(x) is countably compact, and thus C(x) is compact. Indeed, we prove that every sequence in C(x) has a cluster point in it. Take (yj )j in C(x), assume j j j that (xn )n converges to x for every j ∈ N and let yn ∈ φ(xn ) be such that yj is a j j ∈ N cluster point of (yn )n. There exist natural numbers ni ,fori, j , such that

~~≤ j j ··· j ··· ∈ N 1 n1~~

~~1 1 1 1 1 2 2 1 1 x1 ,x2 ,...,x 1− ,x 1 ,...,x 1− ,x 2 ,...,x 2− ,x 1 ,...,x 1− , (3.3) n2 1 n2 n3 1 n2 n3 1 n3 n4 1~~

~~2 2 3 3 1 x 2 ,...,x 2− ,x 3 ,...,x 3− ,x 1 ,... (3.4) n3 n4 1 n3 n4 1 n4 converges to x, and we also have~~

yj ∈ φ(xn) n∈N 3.2 Resolutions and K-analyticity 81 for every j ∈ N. Then (yj )j has a cluster point y ∈ Y . Note that y belongs to C(x) because, if we consider the sequence (zn)n corresponding to (3.3) and (3.4) and deﬁned by

~~1 1 1 1 1 2 2 1 1 y1 ,y2 ,...,y 1− ,y 1 ,...,y 1− ,y 2 ,...,y 2− ,y 1 ,...,y 1− , (3.5) n2 1 n2 n3 1 n2 n3 1 n3 n4 1~~

~~2 2 3 3 1 y 2 ,...,y 2− ,y 3 ,...,y 3− ,y 1 ,... , (3.6) n3 n4 1 n3 n4 1 n4 then zn ∈ φ(xn), and y is a cluster point of (zn)n. Claim 3.1 is proved.~~

Claim 3.2 If G is an open set in Y such that C(x) ⊂ G, there is an open neighborhood V of x such that φ(V) ⊂ G. Indeed, assume the claim fails. Then, for every ∈ N ∈ x ⊂ ∈ n there is xn Vn such that φ(xn) G. Hence we can ﬁnd yn φ(xn) such that yn ∈ Y \ G. Observe that xn → x, and therefore (yn)n has a cluster point y ∈ Y \ G by the assumption, which is a contradiction since by the deﬁnition y ∈ C(x) ⊂ G.

Claim 3.3 If G is an open set in Y such that C(x) ⊂ G, there is an open neighborhood V of x such that φ(V) ⊂ G. Indeed, if O ⊂ Y is open such that C(x) ⊂ O ⊂ O ⊂ G, then applying Claim 3.2 to C(x) and O, one gets an open neighborhood V of x such that φ(V)⊂ O. Finally, φ(V)⊂ O ⊂ G.

Claim 3.4 ψ(x) = C(x), and thus ψ(x) is compact-valued. Indeed, the inclusion C(x) ⊂ ψ(x) is a consequence of the definitions of the sets C(x), ψ(x), and the definition of the cluster point of a sequence. The inclusion C(x) ⊂ ψ(x) follows from the fact that ψ(x) is closed. To prove the converse, fix z ∈ ψ(x). Note that z ∈ C(x). Indeed, for any closed neighborhood U of z in Y and for every n ∈ N ∈ ∩ x ∈ x ∈ ∩ → there is yn U φ(Vn ). Choose xn Vn and yn φ(xn) U. Then xn x, and therefore (yn)n has a cluster point y in Y because of the hypothesis. By the definition, y ∈ C(x), and since U is closed, y ∈ U. Hence z ∈ C(x). We have proved Claim 3.4.

Claim 3.5 ψ : X → 2Y is usco. Indeed, we show that for every open set G ⊃ ψ(x) there is an open neighborhood V of x such that ψ(V)⊂ G.TakeG as above. Since ψ(x) = C(x), we apply Claim 3.3 to get an open neighborhood V of x such that φ(V)⊂ G. Now V is also an open neighborhood of any y ∈ V , and by the deﬁnition we have ψ(y)⊂ φ(V)⊂ G. Hence ψ(V)⊂ G. We have proved the upper semicontinuity of ψ. Finally, observe that by the deﬁnition φ(x) ⊂ ψ(x) for every x ∈ X. The claims together yield the result.

Theorem 3.3 applies for many topological (vector) spaces Y , such as Lindelöf spaces, realcompact spaces, angelic spaces, Banach spaces with the weak topology and others. Theorem 3.3 can also be used to prove the following corollary; see [85, Corollary 4.2]. 82 3 K-analytic and Quasi-Suslin Spaces

Corollary 3.10 Let E be a K-analytic space. Let f : E → Y be a map from E onto a regular topological space whose relatively countably compact subsets are relatively compact. Assume that if (xn)n has a cluster point in E, the sequence (f (xn))n has a cluster point in Y . Then Y is K-analytic. : NN → E = Proof Let θ 2 be an usco map such that E α∈NN θ(α). Deﬁne a map φ : NN → 2Y by φ(α) := f(θ(α)).Themapφ satisﬁes the condition in Theorem N N 3.3. Indeed, if (αn)n is a sequence in N that converges to α ∈ N , then φ(αn) = f(θ(αn)) = f θ(αn) , n n n and n θ(αn) is relatively countably compact. Now we use the condition about the map f , and Theorem 3.3 applies to get an usco map as required. This proves that Y is K-analytic.

~~3.3 Quasi-(LB)-spaces~~

In this section, we discuss a class of lcs E such that E is covered by a resolution consisting of Banach discs. Most results presented here are due to Valdivia [422]. Grothendieck [189] conjectured that the closed graph theorem for a linear map T : E → F holds when the domain space E is ultrabornological (i.e., E is the inductive limit of a family of Banach spaces) and the range space F belongs to a class of lcs that contains the Banach spaces and is stable under countable topological products, countable topological direct sums, separable quotients and closed subspaces. For example, the space of the test functions D(Ω) and the space of distributions D(Ω) belong to this class. Clearly, Grothendieck’s conjecture reduces to the case where E is a Banach space. Slowikowski [382] deﬁned a class of range spaces F that contain D(Ω) and D(Ω) and verify the closed graph theorem when the domain space is a Banach space. Also, Raikov [340] presented a solution for Grothendieck’s conjecture. Valdivia [425] proved that the range class introduced by Raikov coincides with the class given by Slowikowski. De Wilde [110] provided a positive answer for Grothendieck’s conjecture for webbed spaces F ; see also [112]. Valdivia [422] presented another solution for Grothendieck’s conjecture with the class of quasi-(LB)-spaces F ; that is, a subclass of the class considered by Slowikowski. In fact, a quasi-(LB)-space is a strict webbed space, and a strict webbed space is a Slowikowski space. It turns out that for an lcs the classes of quasi-(LB)-spaces and ordered strict webbed spaces are the same [422]. N An lcs E is a quasi-(LB)-space if E admits a resolution {Aα : α ∈ N } consisting of Banach discs [422] (called a quasi-(LB)-representation of E). Quasi-(LB)- spaces enjoy good hereditary properties. We gather some of them. 3.3 Quasi-(LB)-spaces 83

~~Proposition 3.15 If G is a closed subspace of a quasi-(LB)-space E, the space G is a quasi-(LB)-space.~~

~~N Proof Let {Aα : α ∈ N } be a quasi-(LB)-representation of E. The normed space~~

FAα∩G is a normed subspace of the Banach space FAα . Therefore, if (xn)n is a ∈ Cauchy sequence in FAα∩G, and if x is the limit of (xn)n in FAα , then x G by the closedness, and therefore (xn)n converges to x in the Banach space FA ∩G. N α Therefore {Aα ∩ G : α ∈ N } is a quasi-(LB)-representation of G.

Proposition 3.16 Let G be a continuous image of a quasi-(LB)-space F under a linear map T . Then G is a quasi-(LB)-space. In particular, a Hausdorff quotient of a quasi-(LB)-space is a quasi-(LB)-space.

N Proof Let {Aα : α ∈ N } be a quasi-(LB)-representation of F . It is easy to prove := ∩ −1 that Vα FAα T (0) is a closed subspace of the Banach space FAα , and by the construction the normed space GT(Aα) is norm-isomorphic to the quotient space { : ∈ NN} FAα /Vα. Then GT(Aα) is a Banach space, and T(Aα) α is a quasi-(LB)- representation of G.

~~Proposition 3.17 A countable product of quasi-(LB)-spaces is a quasi-(LB)- space.~~

Proof Let J be a countable set, and let (Fj )j∈J be a sequence of quasi-(LB)- j N spaces. Let {Aα : α ∈ N } be a quasi-(LB)-representation of Fj for each j ∈ J . N Let t : N → J × N be a bijection. For αj = (aj,n) ∈ N , j ∈ J ,setat(n) := bn. : NN J → NN { : ∈ } = ∈ NN Deﬁne a bijection T ( ) by T( αj j J ) (bn).Letα and { : ∈ }= −1 := j { : ∈ NN} αj j J T (α). DeﬁneAα j Aαj . Then Aα α is a quasi-(LB)- representation in the product i Fi .

Proposition 3.18 A topological direct sum of a countable family of quasi-(LB)- spaces is a quasi-(LB)-space. := Proof Let E i∈N Fi be the topological direct sum of quasi-(LB)-spaces Fi . { i : ∈ NN} ∈ N Let Aα α be a quasi-(LB) -representation of Fi , i . It is well known that n n the ﬁnite topological product i=1 Fi and the ﬁnite topological direct sum i=1 Fi are isomorphic. Therefore, for each α = (an)n, we note by Proposition 3.17 that

~~:= 1 + 2 +···+ a1 Aα Aα Aα Aα := n { : ∈ NN} is a Banach disc in F i=1 Fi . Clearly, Aα α is a quasi-(LB)-representation of E.~~

Example 1. Every (LF )-space is a quasi-(LB)-space. First note that each Fréchet space E is a quasi-(LB)-space. Indeed, if (Un)n is a decreasing basis of abso- := lutely convex closed neighborhoods of zero in E,thesetsAα n anUn for 84 3 K-analytic and Quasi-Suslin Spaces

N α = (an) ∈ N form a quasi-(LB)-representation of E. Next, since every (LF )- space is isomorphic to a quotient of a countable topological direct sum of Fréchet spaces, it is enough to see that Hausdorff quotients and countable topological direct sums of quasi-(LB)-spaces are quasi-(LB). This we showed above. Example 2. The strong dual of each (LF )-space is a quasi-(LB)-space. Indeed, we know already that the countable topological product of quasi-(LB)-spaces is a quasi-(LB)-space. Next, observe that, if E is a Fréchet space, the strong dual (E ,β(E ,E))of E is a quasi-(LB)-space. Indeed, let (Un)n be a countable basis of absolutely convex closed neighborhoods of zero in E. Set A := U ◦ for α = (a ) ∈ α a1 n N ◦ N N , where U means the polar of U. The family {Aα : α ∈ N } is a quasi-(LB)- representation of (E,β(E,E)). Finally, assume that E is an (LF )-space with a deﬁning sequence (En)n of Fréchet spaces. Note that (E ,σ(E ,E)) is isomorphic to a closed subspace of the product n(En,σ(En,En)). Then (E ,σ(E ,E)) is a quasi-(LB)-space. Since (E,σ(E,E)) and (E,β(E,E)) have the same Banach discs, the conclusion follows.

Proposition 3.19 An inﬁnite-dimensional separable Banach space E admits a N quasi-(LB)-representation {Aα : α ∈ N } of E consisting of absolutely convex compact sets. Hence the unit ball in E cannot be included in any Aα.

Proof By [421, p. 221], the space E is isomorphic to a quotient λ1/F of a Montel echelon space λ1 := λ1(N,A); we refer the reader to Chapter 6 and Chapter 16 of this book for more information about spaces λ1 := λ1(N,A).LetT be a continuous linear map from λ1/F onto E.Let(Un)n be a countable basis of absolutely convex closed neighborhoods of zero in λ1. Then (as is easily seen) the sets Aα := anUn n

N for α = (an) ∈ N form a resolution consisting of absolutely convex bounded and closed sets. Since λ1 is a Montel space (i.e., every bounded and closed set in λ1 is N compact), {Aα : α ∈ N } is a compact resolution in λ1 of absolutely convex sets. N Then {q(Aα) : α ∈ N } is a compact resolution consisting of absolutely convex sets in λ1/F, where q : λ1 → λ1/F is the quotient map. Consequently, N {T(q(Aα)) : α ∈ N } is a compact resolution in E consisting of absolutely convex elements. Since E is inﬁnite-dimensional, the unit closed ball in E is not contained in any set T(q(Aα)).

~~N If α = (an)n ∈ N , k ∈ N, then, as usual, we set α|k := (a1,a2,...,ak). N If m1,m2,...,mk are natural numbers, and {Aα : α ∈ N } is a quasi-(LB)- representation F ,weset~~

~~:= { : ∈ NN | = } Cm1,m2,...,mk Aα α ,α k (m1,m2,...,mk) . (3.7) 3.3 Quasi-(LB)-spaces 85~~

~~For a linear map T : E → F ,weset := −1 Um1,m2,...,mk T (Cm1,m2,...,mk ). Clearly,~~

~~= { : ∈ N} E Um1 m1 and~~

~~= { : ∈ N} Um1,m2,...,mk Um1,m2,...,mk+1 mk+1 for each k ∈ N.IfE is a Baire lcs, there exists a sequence (rn)n in N such that each~~

U r1,r2,...,rk is a neighborhood of zero in E. Also, it is known (see [246, 16.4(7)]) that if C is an absolutely convex neighborhood of zero in an lcs E, and pC is the Minkowski functional of C, the closure C and the interior C0 of C are described by

~~0 C ={x ∈ E : pC(x) ≤ 1},C ={x ∈ E : pC(x) < 1}. The argument of the following technical lemma follows from [422, Proposition 10].~~

Lemma 3.2 Let E be an lcs. Let (Uk)k be a decreasing sequence of absolutely convex sets such that each U k is a neighborhood of zero. For each k ∈ N, let pk be the Minkowski functional of the set U k. Let τ be a locally convex pseudometrizable 0 topology on E deﬁned by the family of seminorms (pn)n. Ifx ∈ (U 1) , then x is the ∈ ∈ N | | limit in τ of a series k αkxk with xk Uk for k , and k αk < 1.

~~Proof Clearly, there exists λ>1 such that λx ∈ U 1. Therefore there exists x1 ∈ U1 and λ − 1 y ∈ U (3.8) 2 22 2 such that~~

~~λx = x1 + y2. (3.9) From (3.8), it follows that there exist~~

~~λ − 1 x ∈ U ,y ∈ U , (3.10) 2 2 3 23 3 such that λ − 1 y = x + y . 2 22 2 3 Then λ − 1 λx = x + x + y . (3.11) 1 22 2 3 86 3 K-analytic and Quasi-Suslin Spaces~~

~~From (3.10), we note that there exist~~

~~λ − 1 x ∈ U ,y ∈ U , 3 3 4 24 4 such that λ − 1 y = x + y . 3 23 3 4 Hence λ − 1 λ − 1 λx = x + x + x + y . 1 22 2 23 3 4~~

~~By an obvious induction, we construct two sequences (xn)n, (yn+1)n, such that for each n ∈ N~~

λ − 1 xn ∈ Un,yn+ ∈ Un+ , (3.12) 1 2n+1 1 and λ − 1 λ − 1 λ − 1 λx = x + x + x +···+ xn + yn+ . 1 22 2 23 3 2n 1 From 1 λ − 1 λ − 1 λ − 1 1 x = x + x + x +···+ xn + yn+ , λ 1 λ22 2 λ23 3 λ2n λ 1 −1 −1 −k we have the lemma with α1 = λ and αk = (λ − 1)λ 2 for k ≥ 2. Indeed, by (3.12), we have

lim yn+1 = 0 n (in τ ), and 1 λ − 1 1 1 1 1 λ − 1 1 λ − 1 |αk| = + + + +··· = + < + = 1. λ λ 22 22 22 λ 2λ λ λ k We shall say that the series k αkxk constructed in Lemma 3.2 is a τ - 0 representation of the point x ∈ (U 1) by the elements of the sequence (Uk)k. N If the sequences α(n) = (α(n)k)k ∈ N , n ∈ N,verifyα(k)|k = α(k + s)|k for each k,s ∈ N, the supremum bk = sup{α(n)k : n = 1, 2,...} is ﬁnite for k ∈ N. N Therefore, β := (bk)k ∈ N and β = sup{α(n) : n ∈ N}. This is applied to prove the following; see [422, Proposition 9].

N Proposition 3.20 Let {Aα : α ∈ N } be a quasi-(LB)-representation of an lcs E. For a sequence (mn)n in N and a neighborhood V of zero in E, there exists k ∈ N 3.3 Quasi-(LB)-spaces 87 such that −1 ⊂ k Cm1,m2,...,mk V.

N Proof Assume the conclusion fails. Then there exists α(k) ∈ N such that α(k)|k = −1 (m1,m2,...,mk) and k Aα(k) V for k ∈ N.Ifβ = sup{α(k) : k ∈ N}, then −1 Aα(k) ⊂ Aβ . Hence, for each k ∈ N we have k Aβ V, contradicting the boundedness of the Banach disc Aβ .

~~:= −1 We supplement Lemma 3.2, where Um1...mk T (Cm1m2...mk ).~~

Proposition 3.21 (Valdivia) Let E and F be lcs. Let T : E → F be a continuous N linear map. Assume that F admits a quasi-(LB)-representation {Aα : α ∈ N } and N that there exists a sequence (rn)n in such that each U r1,r2,...,rk is a neighborhood of the origin in E. Then

~~0 ⊂ (U r1,r2,...,rk ) Ur1,r2,...,rk for each k ∈ N.~~

Proof If V is a closed neighborhood of zero in F , Proposition 3.20 implies that ∈ N −1 ⊂ there exists s such that s Cr1,r2,...,rs V . By the continuity of T , we know = −1 that for the set Ur1,r2,...,rs T (Cr1,r2,...,rs ) we have −1 ⊂ −1 s U r1,r2,...,rs T (V ). (3.13)

Let τ be a locally convex pseudometrizable topology generated by the Minkowski ∈ N functionals of the sets U r1,r2,...,rk for k .From(3.13), it follows that the linear map T : (E, τ) → F is also continuous. By Lemma 3.2,fork ∈ N, each

∈ 0 x (U r1,r2,...,rk ) (3.14) admits a τ -absolutely convex representation ∞ αk+sxk+s s=0 with xk+s ∈ Ur ,r ,...,r + and Σ |αk+s| < 1, and the point x is a limit in τ of the ∞ 1 2 k s series Σs=0αk+sxk+s . The continuity of T yields ∞ T(x)= αk+sT(xk+s). (3.15) s=0 From ∈ = T(xk+s) T(Ur1,r2,...,rk+s ) Cr1,r2,...,rk+s , 88 3 K-analytic and Quasi-Suslin Spaces for s = 0, 1,...,we note that for each s there exists β(s) ∈ NN with

~~β(s)|k+s = (r1,r2,...,rk+s) and T(xk+s) ∈ Aβ(s). For β := sup{β(s) : s ∈ N},wehave~~

T(xk+s) ∈ Aβ for each s ∈ N.AsAβ is a Banach disc and |αk+s| < 1, it follows that in the Banach space generated by Aβ ∞ = ∈ ⊂ αk+sT(xk+s) y Aβ Cr1,r2,...,rk . (3.16) s=0 From (3.15) and (3.16), we note that y = T(x). Therefore ∈ −1 ⊂ −1 = x T (y) T (Cr1,r2,...,rk ) Ur1,r2,...,rk , (3.17) and, using (3.14) and (3.17), we conclude that 0 ⊂ (U r1,r2,...,rk ) Ur1,r2,...,rk (3.18) for each k ∈ N.

~~We shall also need the following useful fact.~~

~~Lemma 3.3 A linear map T : E → F between tvs’s E and F has closed graph if and only if there exists on F a weaker Hausdorff topology ξ such that T : E → (F, ξ) is continuous.~~

Proof Assume the map T : E → F has a closed graph. Let V and W be fundamental systems of balanced neighborhoods of zero in E and F , respectively. The family of balanced sets {T(V)+ W : V ∈ V ,W ∈ W } is a basis of balanced neighborhoods of zero for a vector topology ξ on F weaker than the initial one of F .If y ∈ {T(V)+ W : V ∈ V ,W ∈ W }, (3.19) for each pair (V, W) ∈ V × W there exist xv ∈ V and yw ∈ W such that

~~y = T(xv) + yw. (3.20)~~

~~With the inclusion ⊃ as preorder in V × W for the nets {xv : (V, W) ∈ V × W , ⊃} and {yw : (V, W) ∈ V × W , ⊃},wehave~~

~~lim xv = 0, lim yw = 0. (3.21) (V,W)∈V ×W (V,W)∈V ×W 3.3 Quasi-(LB)-spaces 89~~

~~By (3.20), it follows that~~

~~lim T(xv) = y, (3.22) (V,W)∈V ×W and by (3.21) and (3.22)~~

~~lim (xv,T(xv)) = (0,y). (3.23) (V,W)∈V ×W~~

~~Since the graph of T is closed, y = T(0) by (3.23). Therefore, using (3.19), and since T(0) = 0, we have {T(V)+ W : V ∈ V ,W∈ W }={0}.~~

~~This proves that ξ is Hausdorff. The converse is clear.~~

~~Now we are ready to prove the following interesting closed graph theorem due to Valdivia; see [422, Proposition 11].~~

Theorem 3.4 (Valdivia) Let E and F be lcs’s. Let T : E → F be a linear map with N closed graph. Suppose that F admits a quasi-(LB)-representation {Aα : α ∈ N } N ∈ N and that there exists a sequence (rn)n in such that each U r1,r2,...,rk , k , is a neighborhood of zero in E. Then T is continuous.

Proof Let V be a neighborhood of zero in F . By Proposition 3.20, there exists a positive integer s such that −1 ⊂ s Cr1,r2,...,rs V. By Lemma 3.3, there exists on F a weaker Hausdorff topology ξ such that T : E → N (F, ξ) is continuous. Clearly, {Aα : α ∈ N } is also a quasi-(LB)-representation for (F, ξ). By Proposition 3.21,

~~= −1 Ur1,r2,...,rs T (Cr1,r2,...,rs ) is a neighborhood of zero in E.By~~

~~−1 = −1 −1 ⊂ −1 s Ur1,r2,...,rs s T (Cr1,r2,...,rs ) T (V ) and Proposition 3.21, we obtain the continuity of T : E → F .~~

~~This easily yields the classical Banach closed graph theorem.~~

~~Corollary 3.11 If E is a Baire lcs, F is a metrizable and complete lcs and T : E → F is a linear map with closed graph, then T is continuous.~~

~~We provide another interesting property for quasi-(LB)-spaces; see [422, Proposi- tion 22]. 90 3 K-analytic and Quasi-Suslin Spaces~~

~~Theorem 3.5 (Valdivia) If E is a quasi-(LB)-space, there exists a quasi-(LB)- N representation {Kα : α ∈ N } such that every Banach disc in E is contained in some Kα.~~

{ : ∈ NN} Proof Let Aα α be a quasi-(LB)-representation of E. Deﬁne Ca1,a2,...,an = ∈ NN ∈ N for α (ak) .LetFa1,a2,...,an be the linear span of Ca1,a2,...,an for all n . Set := Fα Fa1,a2,...,an . n

~~Equip the space Fα with the topology ξα having a basis of neighborhoods of zero of the form ∩ −1 (Fα n Ca1,a2,...,an )n.~~

Clearly, ξα is stronger than the original topology induced from E. Moreover, (Fα,ξα) is a Fréchet space; see [422, Proposition 21] for details. This procedure N N provides a family {Fα : α ∈ N } of Fréchet spaces. For α := (an) ∈ N ,set := := ∩ α1 (a2n−1), Kα a2n(Fα1 Ca1,a3,...,a2n−1 ). n

N Then {Kα : α ∈ N } is a quasi-(LB)-representation of E.LetA be a Banach disc in E, and let FA be the Banach space associated with A whose Banach topology is generated by the Minkowski functional norm. Consider the natural continuous N inclusion J : FA → E. We show that there exists β = (bn) ∈ N such that J : FA → N Fβ is continuous and J(A) is bounded in Fβ . This will provide γ ∈ N with A ⊂ Kγ . Set := −1 Ua1,a2,...,an J (Ca1,a2,...,an ) for each n ∈ N and all a1,a2,...,an ∈ N. Since

~~= = FA Ua1 ,Ua1,a2,...,an Ua1,a2,...,an,an+1 a1 an+1~~

N for each n ∈ N and FA is a Banach space, there exists β = (bn) ∈ N such that ∈ N U b1,b2,...,bn is a neighborhood of zero in FA for each n . By Proposition 3.21, 0 ⊂ (U b1,b2,...,bn ) Ub1,b2,...,bn . ∈ ∈ N ∈ ∈ For x FA and k , there exists t>0 such that tx Ub1,b2,...,bk . Hence tJ(x) ∈ ⊂ Cb1,b2,...,bk ,soJ(x) Fb1,b2,...,bk . This proves that J(A) Fβ . Since J : FA → E is continuous, J : FA → Fβ has a closed graph and hence is continuous by Corollary 3.11. Finally, choose a sequence (rn)n in N such that ⊂ ∩ A rk(Fβ Cb1,b2,...,bk ). k 3.4 Suslin schemes 91

~~Set γ = (wn), where~~

~~w2n−1 := bn,w2n := rn, for each n ∈ N. We showed that A ⊂ Kγ .~~

~~We have the following corollary from [422, Corollary 1.6].~~

~~Corollary 3.12 (Valdivia) Every Baire lcs that is a quasi-(LB)-space is a Fréchet space.~~

~~3.4 Suslin schemes~~

This section deals with a very applicable concept, called a Suslin scheme, that provides a powerful tool to obtain several structure theorems for separable metric spaces; see [346] as a good source for this section and the next. The main result of this section, due to Hurewicz, states that every analytic metrizable topological space that is not a σ -compact space contains a closed subspace that is homeomorphic to the space NN. Set

~~N(N) := N0 ∪ Nn, n 0 N where N := ∅.Forσ = (σn) ∈ N , as usual we set~~

~~σ |0 =∅,σ|n := (σ1,σ2,...,σn).~~

~~Assume X is an arbitrary set. A Suslin scheme on X is a map~~

~~N A(.) : N( ) → 2X.~~

If X is a topological space, the Suslin scheme A(.) is said to be open (closed)ifall values A(σ |n) are open (closed) sets for σ |n ∈ N(N). If (X, d) is a metric space, we say that a Suslin scheme A(.) satisﬁes the diameter condition, if for each σ ∈ NN one has

lim diam A(σ |n) = 0. n→∞ | If a Suslin scheme satisfies the diameter condition, n A(σ n) is either the empty set or contains at most one point. Thus, we may define a function ϕ : Z → X by the formula ϕ(σ) = A(σ |n), n 92 3 K-analytic and Quasi-Suslin Spaces where Z = σ ∈ NN : A(σ |n) = ∅ . n The map ϕ is called the map associated to the Suslin scheme A(.). The first result collects some fundamental properties about Suslin schemes.

Proposition 3.22 The map ϕ : Z → (X, d) associated to a Suslin scheme A(.) satisfying the diameter condition has the following properties: (a) ϕ is continuous. N (b) If (X, d) is complete and A(.) is closed, Z is a closed subset of N . N (c) If for each σ ∈ N , k ∈ N, we have {A(σ |n) : n = 0, 1,...,k}=∅and A(σ |k) ⊂ {A(σ |k,q) : q ∈ N}, then Z is a dense subset of NN. (d) If A(.) is open, the map ϕ : Z → ϕ(Z) is open. [ | ]∩[ | ]=∅ = (e) If n A(σ n) n A(τ n) for all σ τ , then ϕ is injective. (f) If A(.) is open and for each σ |n ∈ N(N) the family {A(σ |n, q) : q ∈ N} consists of pairwise disjoint subsets of A(σ |n), then ϕ is a homeomorphism of Z onto ϕ(Z). (g) If A(∅) = X and {A(σ |n, q) : q ∈ N} is a covering of A(σ |n) for each σ |n ∈ N(N), then ϕ(Z) = X.

Proof (a) If σ and τ belong to Z, and σ |n = τ|n, then d(ϕ(σ ), ϕ(τ)) ≤ diam A(σ |n). This combined with the diameter condition implies the continuity of ϕ. N (b) Let (τ[n])n be a sequence in Z converging to σ ∈ N . We may assume that τ[n]|m = σ |m for each n>min N. Therefore

~~ϕ(τ[n]) ∈ A(σ |m) for each n>m. By the diameter condition, we deduce that the sequence (ϕ(τ[n]))n is a Cauchy sequence. The completeness and closedness imply~~

lim ϕ(τ[n]) n→∞ belongs to A(σ |m) for each m ∈ N. Hence σ ∈ Z. Therefore Z is a closed subset of NN. (c) The assumptions from (c) imply that for σ ∈ NN and k ≥ 1 there exists τ ∈ Z such that σ |k = τ|k. Therefore Z is a dense subset of NN. (d) Let σ ∈ Z.Thesets

Cn ={τ ∈ Z : τ|n = σ |n}, for n ≥ 1, determine a neighborhood basis of σ in Z. Since ϕ(Cn) = {A(σ |k) : 0 ≤ k ≤ n} ∩ ϕ(Z), we deduce that φ is open onto the range. The property (e) is clear, and (f) follows directly from the properties (a), (d) and (e). 3.5 Applications of Suslin schemes to separable metrizable spaces 93

~~N If A(.) veriﬁes the property (g), for x ∈ X there exists σ = (σn) ∈ N such that x ∈ A(σ |n) for each n ∈ N. Therefore x = ϕ(σ).~~

~~3.5 Applications of Suslin schemes to separable metrizable spaces~~

We consider a couple of concrete cases; see [346] for results of this section. Case 1. Structure theorems for zero-dimensional metric spaces As usual, |X| denotes the cardinality of a set X. A topological space X is called zero-dimensional if X admits a basis of clopen sets (i.e., sets that are closed and open). It is well known that every nonempty regular topological space X such that |X| ≤ ℵ0 is zero-dimensional. Indeed, since X is Lindelöf, it is normal (see Lemma 6.1). Then, if x ∈ V ⊂ X, where V is open, there exists a continuous function f : X → [0, 1] such that f(x)= 0 and f(X\V)={1}.From|X| ≤ℵ0, it follows that there exists r ∈ I\f(X), where I := [0, 1]. Obviously

− − U = f 1([0,r[) = f 1([0,r]) is a clopen set that contains x and is contained in V . Using the concept of the Suslin scheme, we show some structure theorems for separable metrizable zero-dimensional spaces. In the following, d means a metric compatible with the topology of the space X.

Theorem 3.6 If X is a zero-dimensional separable metrizable space, the space X is homeomorphic to a subset Z of N N . If additionally the space (X, d) is complete, then X is homeomorphic to a closed subset Z of N N .

Proof It is easy to construct a Suslin scheme A(.) on X satisfying the following conditions: (i) A(∅) = X and, for each n ≥ 1 and σ ∈ NN,thesetA(σ |n) is a clopen set with diam A(σ |n) ≤ 2−n−1. (ii) {A(σ |n, q) : q ∈ N} is a partition of A(σ |n) for each σ |n ∈ N(N). Then the map ϕ associated to A(.) is a homeomorphism of Z onto X (compare (f) and (g) from Proposition 3.22). Moreover, by property (b) of Proposition 3.22,if(X, d) is complete, Z is a closed subset of NN.

If all the sets A(σ |n) from the Suslin scheme A(.) are nonvoid, the set Z in Theorem 3.6 is dense in NN. This is the case for the next two theorems. Theorem 3.7 is due to Sierpinski;´ see [346]. 94 3 K-analytic and Quasi-Suslin Spaces

~~Theorem 3.7 (Sierpinski)´ If X is a countable metric space without isolated points, then X is homeomorphic to the space of rational numbers Q.~~

Proof By the remark above, the space X is zero-dimensional. If A is a nonvoid clopen subset of X,thesetA is inﬁnite because X has no isolated points. Set A := {a1,a2,...,}.Forε>0, let A1 be a clopen neighborhood of a1 such that A1 ⊂ A and diam A1 ≤ ε, A1 = A.Nowset

n = min{j ∈ N : aj ∈/ A1}, and pick a clopen neighborhood A2 of a2 such that A2 ⊂ A\A1 and diam A2 ≤ ε, A2 = A\A1. Continuing this way, we construct a sequence of disjoint clopen non- = ≤ ∈ N empty sets (An)n such that A j Aj ,diamAn ε, for each n . Therefore, similar to the preceding case, we can construct a Suslin scheme A(.) on X satisfying both properties considered above and additionally with each set A(σ |n) = ∅ such that: (i) A(∅) = X, and for each n ≥ 1 and σ ∈ NN the set A(σ |n) is a nonvoid clopen set with diam A(σ |n) ≤ 2−n−1. (ii) {A(σ |n, q) : q ∈ N} is a partition of A(σ |n) for each σ |n ∈ N(N). Then, by Proposition 3.22,themapϕ associated to A(.) is a homeomorphism of Z onto X (see properties (f) and (g)), and Z is a dense subset of NN by property (c) of Proposition 3.22. This proves that X is homeomorphic to a dense countable subset of NN, and since NN is homeomorphic to the space of the irrational numbers, X is homeomorphic to a countable dense subset of R. Then X is homeomorphic to Q.

~~Lemma 3.4 leads to a Suslin scheme that will be used in Theorem 3.8.~~

Lemma 3.4 Let (X, d) be a complete metric space such that each compact subset of X has an empty interior. Then, for each nonempty open set A, there exists εA > 0 ⊂ = ∅ ∈ N such that, if A q Bq and diam Bq <εA, then Bq for inﬁnitely many q .

Proof If (X, d) is complete, a subset M ⊂ X is relatively compact if and only if it is totally bounded; that is, for every ε>0 there is a ﬁnite cover of M by sets of diam M ≤ ε. By the assumption on A, we deduce that A is not totally bounded, and the conclusion follows.

~~The following result is due to Alexandrov and Urysohn; see [346].~~

~~Theorem 3.8 (Alexandrov–Urysohn) If (X, d) is a separable, complete and zero- dimensional metric space such that each compact subset of X has an empty interior, then X is homeomorphic to NN.~~

Proof Lemma 3.4 can be used to construct on X a Suslin scheme A(.) such that: (i) A(∅) = X, and, for each n ≥ 1, σ ∈ NN,thesetA(σ |n) is a nonvoid clopen set with diam A(σ |n) ≤ 2−n−1. 3.5 Applications of Suslin schemes to separable metrizable spaces 95

(ii) {A(σ |n, q) : q ∈ N} is a partition of A(σ |n) for each σ |n ∈ N(N). Using properties (b), (c) and (f) of Proposition 3.22, we can see that the map ϕ associated to A(.) is a homeomorphism from NN onto X.

~~Case 2. Structure theorems for separable metric spaces~~

Theorem 3.9 Let (X, d) be a separable metric space. Then X is a continuous open image of a dense subset Z of NN. If (X, d) is complete, the space X is a continuous open image of NN. Hence every Polish space X is a continuous open image of NN.

Proof It is straightforward to construct on (X, d) an open Suslin scheme B(.) such that: (i) B(∅) = X, and for n ≥ 1 and σ ∈ NN one has B(σ|n) = ∅ and diam B(σ|n) ≤ −n−1 2 . | = | = | | ∈ N(N) (ii) B(σ n) q B(σ n, q) q B(σ n, q) for each σ n . Clearly, the map ϕ associated to B(.) coincides with the one that is associated to the Suslin scheme B(.). By conditions (a), (c), (d) and (g) of Proposition 3.22,we deduce that ϕ is continuous and open from Z onto X, where Z is a dense subset of NN. Finally, if (X, d) is complete, then Z = NN by applying (b) of Proposition 3.22 to the Suslin scheme B(.) for which the associated map is ϕ.

~~For another result of this type, we need the following lemma.~~

~~Lemma 3.5 If C is an Fσ -set in a separable metric space X := (X, d), and ε> 0, there exists a sequence (Ck)k of pairwise disjoint Fσ -sets such that Ck ⊂ C, = diam Ck <εand C k Ck.~~

~~Proof By the assumption, there exists a sequence (Gn)n of closed sets in X such = that C n Gn. Then C = G1 ∪ Gn+1 \ Gm . n m≤n~~

Now the conclusion follows from the following two general observations. (a) If D is a closed subset of X, then D and X\D are covered by two countable families, {Dn : n ∈ N} and {En : n ∈ N}, respectively, of closed subsets of X of diameter less than ε. Then D = Dn \ Dm n m

Theorem 3.10 Let (X, d) be a separable metric space. Then X is a continuous injective image of a subset of Z of NN. If (X, d) is complete, then X is a continuous injective image of a closed subset of NN. Hence every Polish space X is a continuous injective image of a closed subset of NN.

Proof Lemma 3.5 applies to get a Suslin scheme C(.) such that: (i) C(∅) = X, and diam C(σ|n) ≤ 2−n−1 for n ≥ 1 and σ ∈ NN. (N) (ii) For σ |n andτ|n in N withσ |n = τ|n,wehaveC(σ|n) ∩ C(τ|n) =∅. | = | = | | ∈ N(N) (iii) C(σ n) q C(σ n, q) q C(σ n, q) for each σ n . The map ϕ associated to C(.) coincides with the map associated to the Suslin scheme C(.). By conditions (a), (e) and (g) of Proposition 3.22, we conclude that ϕ is an injective and continuous map from Z onto X, where Z is a subset of NN. If (X, d) is complete, Z is closed by (b) of Proposition 3.22 applied to the Suslin scheme C(.).

A sequence (An)n of subsets of a topological space X converges to an element a ∈ X if the ﬁlter base {Ap : p ≥ n} converges to a; we will write a = limn→∞ An. We need a couple of additional lemmas; see [346].

~~Lemma 3.6 Let H(.)be an open Suslin scheme on a complete metric space (X, d) such that for each (σ |n) ∈ N(N)~~

~~− − diam H(σ|n + 1) ≤ 2 n 1, (3.24)~~

~~H(σ|n + 1) ⊂ H(σ|n), (3.25) and H(.) veriﬁes the condition~~

x(σ|n) := lim H(σ|n, q) ∈ H(σ|n)\ H(σ|n, q), (3.26) q→∞ q and if τ|n = σ |n, then H(σ|n) ∩ H(τ|n) =∅. (3.27) Then the set D := {x(σ|n)˙ : (σ |n) ∈ N(N)} is homeomorphic to the topological space Q of the rational numbers, and D\D is homeomorphic to NN.

Proof From (3.26), it follows that x(σ|n) = limq→∞ x(σ|n, q) and x(σ|n) = x(σ|n, q). Theorem 3.7 is applied to deduce that D is homeomorphic to Q. 3.5 Applications of Suslin schemes to separable metrizable spaces 97

From (3.24), (3.25), (3.26) and (3.27), we deduce that for each σ ∈ NN the sequence (x(σ |n))n is a Cauchy sequence whose limit x(σ) veriﬁes {x(σ)}= lim x(σ|n) = H(σ|n) = H(σ|n), (3.28) n→∞ n n

N and ifτ ∈ N \{σ}, then x(σ) = x(τ). Therefore, the equality x(σ) = limn→∞ x(σ|n) deﬁnes a bijection x(.) from NN onto {x(σ) : σ ∈ NN}. From (3.24) and (3.25), we derive that the map x(.) is continuous. The formulas (3.25) and (3.27) imply that x(τ) ∈ H(σ|n) holds for a point τ ∈ NN if and only if τ|n = σ |n. Therefore the map x(.) is open because if

N An := {τ ∈ N : τ|n = σ |n}, then {An : n ∈ N} is a neighborhood basis of σ , and also {H(σ|n) : n ∈ N} is a neighborhood basis of x(σ). Therefore x(.) is a homeomorphism from NN onto {x(σ) : σ ∈ NN}.Ift ∈ D\D, N there exist two sequences (σ (m))m and (n(m))m,inN and N, respectively, such that limm→∞ x(σ(m)|n(m)) = t. Let

~~N Pq : N → N be the projection deﬁned by~~

~~Pq ((a1,a2,...,an,...))= aq .~~

~~From (3.26), it follows that the set {n(m), m ∈ N} is unbounded, and each set~~

~~{Pq (σ (m)|n(m)), m ∈ N,n(m)≥ q} is bounded. The diagonal method, taking an adequate sequence if necessary, enables us to assume that~~

~~Pq (σ (m)) = βq if m ≥ q.Ifβ = (β1,β2,...), from (3.24) and (3.25) we conclude that~~

~~x(β) = lim x(β|m) = lim x(σ(m)|n(m)) = t. m→∞ m→∞~~

~~This proves that D\D ⊂ x(NN). Conversely, if σ ∈ NN, then by (3.28)wehave x(σ) ∈ D. Since {x(σ)}= H(σ|n), n (3.26) and (3.27) ensure that for each n ∈ N and τ ∈ NN\{σ } one has~~

~~x(σ|n) = x(σ) = x(τ|n). 98 3 K-analytic and Quasi-Suslin Spaces~~

~~This proves that x(NN) ⊂ D\D.~~

~~A topological space X is called analytic if X is a continuous image of a Polish space (or equivalently of the space NN). Clearly, every analytic space is K-analytic.~~

Lemma 3.7 Let X be a topological space. Let Y be an analytic subspace of X that is not an Fσ -subset of X. Then there exist a nonempty Polish space T and a X continuous map θ : T → Y such that θ(G) \Y = ∅ for each nonempty open subset G of T .

Proof Since Y is analytic, there exist a Polish space S and a continuous surjection θ0 of S onto Y .LetB be the family of all open subsets of S such that B ∈ B if there exists in X an Fσ -subset A such that

~~θ0(B) ⊂ A ⊂ Y.~~

~~Let G0 be the union of the sets of the family B. Clearly, G0 is the union of a countable family of elements of B. This implies that G0 ∈ B, and then there exists in X an Fσ -subset A0 such that~~

~~θ0(G0) ⊂ A0 ⊂ Y.~~

= := \ −1 TheassumptiononY ensures A0 Y .ThesetT S θ0 (A0) is a nonvoid Gδ- subset of S, and then T is a Polish space. If for a nonvoid open subset G of T one X has θ0(G) ⊂ Y , then G0 ∪G ∈ B, a contradiction with the deﬁnition of G0. Hence

~~X θ0(G) \Y = ∅.~~

~~To complete the proof, we deﬁne θ as the map θ0 restricted to T .~~

~~We also need the following technical lemma.~~

Lemma 3.8 Let X be a metric space. Let Y be an analytic subspace of X that is not an Fσ -subset. Let T and θ be chosen as in Lemma 3.7. Then there exist two open Suslin schemes, G(.) in T and H(.) in X such that for each σ ∈ NN and each n ∈ N

− − diam G(σ |n + 1) ≤ 2 n 1, (3.29) − − diam H(σ|n + 1) ≤ 2 n 1, (3.30) G(σ |n + 1) ⊂ G(σ |n), (3.31) H(σ|n + 1) ⊂ H(σ|n), (3.32) θ(G(σ|n)) ⊂ H(σ|n). (3.33) 3.5 Applications of Suslin schemes to separable metrizable spaces 99

~~Furthermore, H(.) veriﬁes the convergence condition~~

~~x(σ|n) := lim H(σ|n, q) ∈{θ(G(σ|n))\Y }\ H(σ|n, q), (3.34) q→∞ q and if σ |n = τ|n, then H(σ|n) ∩ H(τ|n) =∅. (3.35)~~

~~Proof The construction will be provided by the induction. Step 1. By Lemma 3.7, for the sets G(∅) := T and H(∅) := X there exists a point x(∅) such that x(∅) ∈ θ(G(∅))\Y. (3.36)~~

~~From (3.36), there exists a sequence (z(n))n in θ(G(∅)) such that~~

~~x(∅) = lim z(n). n→∞~~

~~Then there exists a sequence (H (n))n of open neighborhoods of the points z(n), n ∈ N, with pairwise disjoint closures contained in~~

~~X\{x(∅)}=H(∅)\{x(∅)}, and such that for each n ∈ N − diam H(n)≤ 2 1, (3.37) and the sequence (H (n))n converges to x(∅); that is,~~

~~x(∅) = lim H(n). (3.38) n→∞ Then clearly~~

~~x(∅) ∈{θ(G(∅))\Y }\ H(q). (3.39) q~~

~~The continuity of θ enables us to determine a sequence (G(n))n of open subsets of T such that for each n ∈ N − diam G(n) ≤ 2 1, (3.40) θ(G(n))⊂ H(n). (3.41)~~

~~Lemma 3.7 implies that θ(G(n)) \Y = ∅, and therefore we may choose~~

~~x(n) ∈ θ(G(n))\Y (3.42) for each n ∈ N. Then, from (3.38) and (3.41) it follows that~~

~~x(∅) = lim x(n). (3.43) n→∞ 100 3 K-analytic and Quasi-Suslin Spaces~~

Step 2. Assume that for each n ∈ N we already constructed the sets G(n) and H(n)and the point x(n) verifying (3.41) and (3.42). Therefore there exists an injective sequence (z(n, p))p in θ(G(n)) such that x(n) = limp→∞ z(n, p). We may determine a sequence (H (n, p))p of open neighborhoods of the points z(n, p), p ∈ N, with the pairwise disjoint closures contained in H(n)\{x(n)} and such that for each p ∈ N − diam H(n,p)≤ 2 2 (3.44) and the sequence (H (n, p))p veriﬁes the convergence condition

~~x(n) = lim H(n,p). (3.45) p→∞~~

~~It is obvious that~~

~~x(n) ∈{θ(G(n))\Y }\ H(n,p). (3.46) q~~

~~The continuity of θ provides a sequence (G(n, p))p of open subsets of G(n) such that for each p ∈ N − diam G(n, p) ≤ 2 2, (3.47) θ(G(n,p))⊂ H(n,p). (3.48)~~

~~Lemma 3.7 implies that θ(G(n,p)) \Y = ∅, and therefore we may choose~~

~~x(n,p) ∈ θ(G(n))\Y (3.49) for each n, p ∈ N. Then, from (3.45), (3.48) and (3.49) it follows that~~

~~x(n) = lim x(n,p), (3.50) p→∞ and this completes the second step of the induction process. Now it is obvious how to complete the induction process.~~

~~We are ready to prove the following theorem.~~

Theorem 3.11 Let X be a complete metric space. Let Y be an analytic subspace of X that is not an Fσ -subset. Then there exists a closed set F ⊂ X such that: (a) P = F ∩ Y is homeomorphic to NN. (b) Q = F \Y is homeomorphic to Q. (c) Q = F .

Proof Let T and θ be chosen as in Lemma 3.7.LetG(.) and H(.) be the Suslin schemes provided in Lemma 3.8, and let x(σ|n) = limq→∞ H(σ|n, q);see(3.34). From (3.43), (3.50) and Theorem 3.7, it follows that the set Q := {x(σ|n) : n ∈ N(N)} is homeomorphic to the space Q of rational numbers. By (3.36) and (3.42), we have Q ⊂ X\Y . 3.6 Calbrix–Hurewicz theorem 101

Lemma 3.6 implies that the map x(.) : NN → Q\Q given by x(σ) = limn→∞ x(σ|n) is a homeomorphism. Now we prove that Q\Q ⊂ Y . This will show that Q ∩ Y = Q\Q, and the theorem will be proved. If y ∈ Q\Q = ϕ(NN), there exists σ ∈ NN with

~~y = ϕ(σ) = lim x(σ|n). n→∞ By (3.34), we note that x(σ|n) ∈ θ(G(σ|n).~~

~~Therefore there exists tn ∈ G(σ |n) such that −n d(x(σ|n), θ(tn)) ≤ 2 , and we have~~

~~y = lim θ(tn). n→∞~~

~~Clearly, the sequence (tn)n veriﬁes the Cauchy condition (see (3.29) and (3.31)). By the assumption, the space T is Polish, so there exists t ∈ T such that t = limn→∞ tn. By the continuity,~~

~~y = lim θ(tn) = θ(t)∈ θ(T)⊂ Y. n→∞~~

~~We complete this section with the following result of Hurewicz.~~

~~Theorem 3.12 (Hurewicz) If Y is an analytic metrizable space that is not σ - compact, then Y contains NN as a closed subset.~~

Proof Clearly, Y has a countable base. Hence Y is a normal topological space. Therefore Y may be embedded in a topological space X homeomorphic to a product of a countable family of intervals [0, 1]. Since Y is not a σ -compact space, it is not an Fσ -space in X, and the conclusion follows from Theorem 3.11.

~~3.6 Calbrix–Hurewicz theorem~~

The aim of this section is to present an extension of Theorem 3.12. We show a theorem due to Calbrix [77] that in this particular case states that, if X is a regular analytic topological space that is not σ -compact, then X contains a closed subset that is homeomorphic with the space NN. We start with a couple of additional lemmas; see [346]. The ﬁrst one is similar to Lemma 3.7. Recall that a topological space X is hereditarily Lindelöf if each subspace of X is Lindelöf. 102 3 K-analytic and Quasi-Suslin Spaces

Lemma 3.9 Let X be a hereditarily Lindelöf subspace of a topological space Y such that X is not contained in a countable union of compact subsets of Y . Then X has a nonvoid closed subset F such that each nonvoid open subset of F is not contained in a countable union of compact subsets of Y .

Proof Let B be the union of all open subsets of X such that each of them is contained in a countable union of compact subsets of Y . Since B is a Lindelöf space, it is the maximal open subset of X contained in a countable union of compact subsets of Y . Then F := X\B is nonvoid and closed, and veriﬁes the lemma.

For a K-analytic space X,letK : NN → 2X be an usco compact-valued map. N We recall again our notation. For σ = (σ1,σ2,...,σn,...)∈ N ,setσ |0 := ∅ and σ |n := (σ1,σ2,...,σn) for n ∈ N. Deﬁne N I(σ|n) ={τ ∈ N : τ|n = σ |n}, (3.51) where N I(σ|0) = I(∅) = N . (3.52) The sets I(σ|n) compose a basis of neighborhoods of σ in NN.ThemapM(.) : N(N) → 2X, called the Suslin scheme associated to the usco map K, is deﬁned by M(σ|n) := K(I(σ|n)). Clearly,

~~M(σ|0) = M(∅) = X (3.53) and~~

N M(σ|n) = {K(τ) : τ ∈ N ,τ|n = σ |n}. (3.54) = | From the upper semicontinuity of K, we deduce that K(σ) n M(σ n). If, moreover, X is regular, then K(σ) is the intersection of its closed neighborhoods. There- fore, if X is a regular K-analytic space,

~~K(σ) =∩{M(σ|n) : n ∈ N}. (3.55)~~

~~We also need the following lemma.~~

Lemma 3.10 Let X be a K-analytic space. Let M(.) be a Suslin scheme associated with an usco map K : NN → 2X. Let B be a family of subsets of X satisfying the nonvoid ﬁnite intersection property. If there exists σ ∈ NN such that for each n ∈ N there exists Bn ∈ B with Bn ⊂ M(σ|n), then X {B : B ∈ B} ∩ K(σ) = ∅. (3.56)

~~Moreover, if X is a topological subspace of Y , we have X Y {B : B ∈ B}= {B : B ∈ B}. (3.57) 3.6 Calbrix–Hurewicz theorem 103~~

X Proof If [ {B : B ∈ B}] ∩ K(σ) =∅, the compactness of K(σ) yields a ﬁnite X family B0 ⊂ B such that K(σ) ⊂ {X\B : B ∈ B0}. By the upper semicontinuity of K, there exists n ∈ N such that

~~X K(I(σ|n)) = M(σ|n) ⊂ {X\B : B ∈ B0}.~~

X Since Bn ⊂ M(σ|n),wehaveBn ⊂ {X\B : B ∈ B0}, contradicting the ﬁnite intersection property of B. This proves (3.56). Y To prove (3.57), let y ∈ {B : B ∈ B}, and let C be the family of all intersections of neighborhoods of y in Y with sets of B. Applying (3.56) to the family X Y C , we deduce that [ {C : C ∈ C }] ∩ K(σ) = ∅. Since {C : C ∈ C }={y}, y ∈ K(σ) ⊂ X. This shows (3.57).

Lemma 3.11 Let K : NN → 2X be an usco map of a regular K-analytic space X, and let M(.) be a Suslin scheme associated to K. Assume that for each σ ∈ NN there exists a sequence (A(σ |n))n of subsets of X such that

A(σ |n) ⊂ M(σ|n) (3.58) and A(σ |n + 1) ⊂ A(σ |n). (3.59) := | Then A(σ ) n A(σ n) is compact, and if U is a neighborhood of A(σ ) there exists m ∈ N such that A(σ |m) ⊂ U. (3.60) Therefore, if for each σ ∈ NN and n ∈ N the sets A(σ |n) satisfy (3.58) and (3.59), then

~~N A := {A(σ ) : σ ∈ N } (3.61) is K-analytic. The map P : NN → 2A deﬁned by P(σ):= A(σ ) is usco, P(NN) = A and P(σ)⊂ K(σ) for each σ ∈ NN.~~

Proof From (3.55) follows the compactness of A(σ ).IfU is an open set that contains A(σ ) and A(σ |n) ∩ (X\U) = ∅ for each n ∈ N, then applying (3.56)from Lemma 3.10, we note A(σ |n) ∩ (X\U)X ∩ K(σ) = ∅. n On the other hand, A(σ |n) ∩ (X\U)X ∩ K(σ) ⊂ A(σ ) ∩ (X\U)=∅. n 104 3 K-analytic and Quasi-Suslin Spaces

~~This contradiction yields m ∈ N such that (3.60) holds. Then the K-analyticity of A and the desired properties of the map P are clear.~~

~~The next lemma is related to Lemma 3.9. The proof is given for the sake of completeness.~~

Lemma 3.12 Let X be a hereditarily Lindelöf subspace of Y such that X is not a σ -compact subset of Y . Then X has a nonvoid closed subset F such that for each nonvoid open subset A of F the set A is not a σ -compact subset of Y .

Proof Let B be the family of all open subsets of X such that C ∈ B if and only if C is a σ -compact subset of Y . Set B := {C : C ∈ B}.AsB is Lindelöf, then B is a σ -compact subset of Y . Therefore B is the maximal open subset of X such that B is a σ -compact subset of Y . Clearly, the nonvoid closed subset F = X\B is as required.

Lemma 3.13 Let K : NN → 2X be an usco map for a K-analytic regular hereditarily Lindelöf space X, and let M(.) be a Suslin scheme associated to K. Assume that X is a subspace of the hereditarily paracompact topological space Y , and let Y A be a subset of X such that A (the closure in Y ) is not a σ -compact subset of Y . X If s ∈ N(N) such that A ⊂ M(s) and G is an open subset of X such that A ⊂ G, then there exist three sequences (An)n, (Gn)n and (kn)n such that each kn ∈ N and for each n ∈ N Y (1) An ⊂ A such that An is not a σ -compact subset of Y , (2) Gn ⊂ G is open, (3) An ⊂ M((s, kn)), X (4) An ⊂ Gn, Y Y (5) Gn ∩ Gm =∅, if n = m, Y (6) {Gn : n ∈ N} is a locally ﬁnite family in Y .

Proof Changing A by a suitable subset if necessary, we can use Lemma 3.12 to Y assume that for each nonvoid open subset B of A the set B is not a σ -compact Y subset of Y . Hence A is not a compact subspace of Y . Therefore, there exists an Y open covering {L : j ∈ J } of A such that no finite subfamily of {L : j ∈ J } j j Y := { : ∈ } covers A . Set L Li j J . Then, since L is paracompact, there exists { : ∈ } { : ∈ } a locally finite open covering Li i I of L that refines Lj j J (i.e., for ∈ ∈ ⊂ each Li , i I , there exists j J such that Li Lj ). Hence, no finite subfamily { : ∈ } Y ∈ ∈ ∈ of Li i I covers A .Ifx1 A, there exists i1 I such that x1 Li1 , and we ⊂ Y ⊂ choose an open neighborhood G1 of x1 in X such that G1 G and G1 Li1 . As { : ∈ } Y ∈ Y \ no finite subfamily of Li i I covers A , there exists z2 A Li1 , and therefore 3.6 Calbrix–Hurewicz theorem 105

~~∈ \{ } ∈ Y ∩ we may ﬁnd i2 I i1 such that z2 A Li2 . Since~~

∈ Y ∩[ \ Y ] z2 A Li2 G1 , there exists ∈ ∩[ \ Y ] x2 A Li2 G1 . Y Then there exists an open neighborhood G2 of x2 in X such that G2 ⊂ G and G2 ⊂ \ Y Li2 G1 . Choose ∈ Y \{ ∪ } z3 A Li1 Li2 ∈ \{ } ∈ Y ∩ and i3 I i1,i2 such that z3 A Li3 . Since ∈ Y ∩ \{ Y ∪ Y } z3 A Li3 G1 G2 , there exists ∈ ∩ \{ Y ∪ Y } x3 A Li3 G1 G2 .

~~Therefore there exists an open neighborhood G3 of x3 in X such that G3 ⊂ G and~~

~~Y ⊂ \{ Y ∪ Y } G3 Li3 G1 G2 .~~

Y \{ ∪ ∪ }=∅ Since A Li1 Li2 Li3 , we continue this process to determine a sequence (xn)n in A, a sequence (Gn)n such that each Gn is an open neighborhood of xn in = X contained in G, and a sequence (Lin )n such that for m n = Lim Lin (3.62) and Y ⊂ \{ Y ∪ Y ∪···∪ Y } Gn Lin G1 G2 Gn−1 . Y ⊂ ∈ N { Y : By (3.62), and since Gn Lin for each n , we deduce that the family Gn n = 1, 2,...} is locally ﬁnite. X For each n ∈ N,letHn be a neighborhood of xn in X such that Hn ⊂ Gn. The ∩ Y assumption from the beginning of the proof implies that Hn A is not a σ -compact ⊂ := ∩ = [ ∩ ∩ ] subset of Y . Since A M(s) k M(s,k) and Hn A k Hn A M(s,k) , there exists kn such that the closure (in Y )ofthesetAn := Hn ∩ A ∩ M(s,kn) is not a σ -compact subset of Y . We constructed the sequences (An)n, (Gn)n and (kn)n as required.

~~Lemma 3.13 enables us to construct special K-analytic subspaces in certain K-analytic spaces. 106 3 K-analytic and Quasi-Suslin Spaces~~

Lemma 3.14 Let X be a regular hereditarily Lindelöf K-analytic space that is a subspace of a hereditarily paracompact space Y . If X is not a σ -compact subset of Y , there exists a K-analytic subspace A ⊂ X closed in Y and a continuous map f : A → NN such that the map f −1 : NN → 2A is an usco map, and f −1(NN) = A. If additionally X is analytic, f is a homeomorphism.

Proof Let K be an usco map defining the K-analyticity of X.LetM(.) be a Suslin scheme associated with K. Applying Lemma 3.13,forA = G = M(∅) = X we obtain Y (1 ) a sequence (A(σ1))σ1 such that each A(σ1) is a subset of A with A(σ1) not a σ -compact subset of Y ; and := (2 ) a sequence (G(σ1))σ1 such that each G(σ1) is open in G X; ⊂ X ⊂ (3 ) a sequence (kσ1 )σ1 of natural numbers such that A(σ1) M(kσ1 ), A(σ1) ∈ N { Y = } G(σ1) for each σ1 , and that the family Gσ1 ,σ1 1, 2,... is locally finite in Y of pairwise disjoint subsets. = = = Again applying Lemma 3.13 with A A(σ1), G G(σ1) and M(s) M(kσ1 ), we obtain (1 ) a sequence (A(σ1,σ2))σ2 such that each A(σ1,σ2) is a subset of A(σ1) with Y A(σ11,σ2)) and not σ -compact in Y ; (2 ) a sequence (G(σ1,σ2))σ2 such that each G(σ1,σ2) is an open subset of G(σ1); and ⊂ (3 ) a sequence of natural numbers (kσ1σ2 )σ2 with A(σ1,σ2) M(kσ1 ,kσ1σ2 ) for each σ2 ∈ N and X A(σ1,σ2) ⊂ G(σ1,σ2) (3.63) and such that Y {G(σ1,σ2) ,σ2 = 1, 2,...} (3.64) is a locally finite family in Y of pairwise disjoint subsets. Then the family

Y 2 {G(σ1,σ2) ,(σ1,σ2) ∈ N } is also locally ﬁnite in Y , and the sets are pairwise disjoint. Continuing this way, we obtain on X (1) a Suslin scheme A(.), where for each s ∈ NN, p ∈ N,thesetA(s, p) is a subset of A(s) such that A(s, p) is not a σ -compact subset of Y ; (2) a Suslin scheme G(.), where each G(s, p) is an open subset of G(s); and (3) a Suslin scheme k(.), where k(σ1σ2 ...σn) := {kσ σ ...σ }, k(∅) = N, and such N 1 2 n that for each σ = (σi) ∈ N , n ∈ N,thesets

~~A(σ |n) = A(σ1σ2 ...σn), the open sets~~

G(σ |n) = G(σ1σ2 ...σn) 3.6 Calbrix–Hurewicz theorem 107 and the sets of the Suslin scheme M(.) | | | = M(k(σ 1), k(σ 2), . . . , k(σ n)) M(kσ1 ,kσ1σ2 ,...,kσ1σ2...σn ) satisfy A(σ |n) ⊂ M(k(σ|1), k(σ |2),...,k(σ|n)), (3.65)

~~X A(σ |n) ⊂ G(σ |n), (3.66) and for each n ∈ N the family~~

~~Y N {G(σ |n) ,σ ∈ N } (3.67) is locally ﬁnite in Y with pairwise disjoint sets. For each σ ∈ NN,set X A(σ ) := A(σ |n) n and~~

A := {A(σ ) : σ ∈ NN}. From (3.65) and Lemma 3.11, we deduce that A is K-analytic under the mapping P(.): NN → A deﬁned by P(σ)= A(σ ). The formulas (3.66) and (3.67)implyA(σ ) ∩ A(τ) =∅for σ = τ . Therefore, we may deﬁne a map f : A → NN such that f(x)= σ if x ∈ P(σ). The map f is continuous since from (3.66) and 3.67) we note

~~N f(A∩ G(σ1σ2 ...σn)) ⊂{τ ∈ N : τ|n = σ |n}.~~

~~Indeed, if x ∈ A ∩ G(σ1σ2 ...σn) and f(x)= τ , then~~

~~X x ∈ P(τ)= A(τ) ⊂ A(τ|n) ⊂ G(τ1τ2 ...τn).~~

On the other hand, G(τ1τ2 ...τn) ∩ G(σ1σ2 ...σn) = ∅ if and only if τ|n = σ |n. This proves f(x)∈{τ ∈ NN : τ|n = σ |n}. Clearly, P = f −1. Finally, we prove that A is a closed subset of Y. From (3.57) in Lemma 3.10, it follows that X Y A(σ ) = A(σ |n) = A(σ |n) . n n Then A = {A(σ ) : σ ∈ NN} is closed in Y since from (3.66) and (3.67) it follows { : ∈ NN} = | Y that the family A(σ ) σ is locally ﬁnite in Y and each A(σ ) n A(σ n) is closed in Y .IfX is analytic, there exists a compact-valued usco map K : NN → 2X such that each image K(σ),forσ ∈ NN, is a unitary set. Consequently, the 108 3 K-analytic and Quasi-Suslin Spaces nonvoid set P(σ)is unitary. Now the map f : A → NN such that f(x)= σ if {x}= P(σ)is a homeomorphism.

Deﬁnition 3.1 Let A and B be two topological spaces. B is a proper image of A if there exists a continuous map f : A → B such that (1) f(A)= B, (2) for each y ∈ B,thesetf −1(y) is a compact subset of A, (3) the map P : B → 2A deﬁned by P(y)= f −1(y) is usco.

~~Now we are ready to prove the main result; see [346].~~

Theorem 3.13 (Calbrix–Hurewicz) Let X be a regular hereditarily Lindelöf K- analytic space that is a subspace of the hereditarily paracompact topological space Y . Then X is not a σ -compact subset of Y if and only if there exists in X a subspace A that is closed in Y and such that NN is a proper image of A. In particular, if X is a regular and analytic subspace of a hereditarily paracompact topological space Y , then X is not a σ -compact subset of Y if and only if there exists in X a subspace A, closed in Y , such that NN is homeomorphic to A.

~~Proof One part of the proof follows directly from Lemma 3.14. The converse follows from the known fact stating that NN is not a σ -compact space.~~

~~It is well known that a regular and (hereditarily) Lindelöf topological space is (hereditarily) paracompact; see [146]. By Theorem 3.13, we note the following.~~

~~Corollary 3.13 A regular analytic space X is not σ -compact if and only if there exists in X a closed subspace homeomorphic to NN. Chapter 4 Web-Compact Spaces and Angelic Theorems~~

Abstract This chapter deals with the class of angelic spaces, introduced by Frem- lin, for which several variants of compactness coincide. A remarkable paper of Orihuela introduces a large class of topological spaces X (under the name web- compact) for which the space Cp(X) is angelic. Orihuela’s theorem covers many already known partial results providing Eberlein–Šmulian-type results. Following Orihuela, we show that Cp(X) is angelic if X is web-compact. This yields, in particular, Talagrand’s result stating that for a compact space X the space Cp(X) is K-analytic if and only if C(X) is weakly K-analytic. We present some quantitative versions of Grothendieck’s characterization of the weak compactness for spaces C(X) (for compact Hausdorff spaces X) and quantitative versions of the classical Eberlein–Grothendieck and Krein–Šmulian theorems.

~~4.1 Angelic lemma and angelicity~~

In 1940, Šmulian proved that weakly relatively compact sets in a Banach space E are weakly relatively sequentially compact, and if (E,σ(E,E))is separable, then A ⊂ E is weakly relatively countably compact if and only if A is weakly relatively sequentially compact. Dieudonné and Schwartz extended this result to locally convex spaces admitting a weaker metrizable topology. Eberlein showed that weakly relatively countably compact sets are weakly relatively compact for a Banach space E. This result has been generalized by Grothendieck for lcs’s that are quasicomplete in the Mackey topology μ(E, E);see[165]. Finally, Fremlin’s approach to angelic spaces led to presenting all mentioned results in a much easier form; see [165] and [320], [82]. For the deﬁnition of an angelic space, we refer to Chapter 3. In this section, we recall elementary facts about angelic spaces, Fremlin’s angelic lemma and some applications. The following result, due to Fremlin (also called an angelic lemma), will be used several times in the book; see [165].

Theorem 4.1 Let X and Y be topological spaces, where X is regular, and let Φ : X → Y be an injective and continuous map. If A ⊂ X is relatively countably compact, and if for each B ⊂ Φ(A) and y ∈ B there exists a sequence (yn)n in B converging to y, then Φ(A) is closed and Φ|A is a homeomorphism.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 109 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_4, © Springer Science+Business Media, LLC 2011 110 4 Web-Compact Spaces and Angelic Theorems

Proof If D ⊂ A and y ∈ Φ(D), there exists a sequence (dn)n in D such that (Φ(dn))n converges to y.Letx be a cluster point of (dn)n. Since y = Φ(x), y ∈ Φ(D). Hence Φ(D) = Φ(D). This shows also that Φ(A) is closed in Y .Let C be a closed subset of A.Ifx ∈ A \ C, and if V is an open neighborhood of x such that

~~C ⊂ Bx := X \ V, then C ⊂ Dx and x/∈ Dx, where Dx := Bx ∩ A. Therefore C = {Dx : x ∈ A \ C} and Φ(C) = {Φ(Dx) : x ∈ A \ C}= {Φ(Dx) : x ∈ A \ C}.~~

~~Hence Φ(C) is closed and Φ|A is a homeomorphism.~~

Corollary 4.1 Let Φ : X → Y be a continuous injective map deﬁned on a regular topological space X with values in a compact space Y . If for each relatively countably compact subset A of X and for each b ∈ Φ(A) there exists a sequence (an)n in A such that b = limn Φ(an), the space X is angelic.

~~Theorem 4.1 yields the following useful property for angelic spaces; see [165].~~

~~Corollary 4.2 If E is an angelic space, then E endowed with any stronger regular topology than the original one is also angelic.~~

~~Theorem 4.1 is applied to show the following result of Šmulian; see [165, Theo- rem 3.2].~~

Theorem 4.2 Let E be an lcs such that (E,σ(E,E)) is separable. Then, for the space (E, σ (E, E)), the following conditions hold: (i)(Relatively) compact,(relatively) countably compact and (relatively) sequentially compact sets are the same. (ii) If A is relatively (countably) compact and x ∈ A, there exists a sequence in A converging to x. (iii) Every relatively countably compact set that is sequentially closed is compact.

~~We will also need the following fact due to Fremlin; see [165, Theorem 3.5].~~

~~Theorem 4.3 If Cp(X) is angelic, then Cp(X, Z) is angelic for any metric space Z. 4.2 Orihuela’s angelic theorem 111~~

~~4.2 Orihuela’s angelic theorem~~

~~Let Z be a topological space. Let X be a set, and let A ⊂ ZX be a set of functions. X and A aresaidtohavetheinterchangeable double limit property (in Z)if~~

lim lim fm(xn) = lim lim fm(xn) m→∞ n→∞ n→∞ m→∞ for each sequence (xn)n in X and each sequence (fm)n in A if all the limits involved exist. N Let Σ be a subset of N . Assume that there exists a family {Aα : α ∈ Σ} of nonvoid subsets of a set Y covering Y .Forα = (ak) ∈ Σ, as usual set := { : = ∈ = = } Ca1,a2,...,ak Aβ β (bk) Σ, bj aj ,j 1, 2,...,k for every α = (ak) ∈ Σ. We start with the following fundamental result due to Orihuela [320].

Theorem 4.4 (Orihuela) Let (Z, d) be a compact metric space. Let K be a set of functions from Y into Z. Suppose that for each α = (am) ∈ Σ, for each increasing ∈ ∈ N sequence of natural numbers (mp)p, and for each yp Ca1a2...amp , p , one has that each sequence (fn)n in K satisﬁes the interchangeable double limit property Y with (yp). Then, for every f in the closure of K in Z , there is a sequence (gn)n in K converging pointwise to f .

~~Proof The proof will be presented in three steps. Y Step 1. If gi ∈ Z ,1≤ i ≤ m, and D ⊂ Y , then~~

{(g1(x), g2(x), . . . , gm(x)) : x ∈ D} is a subset of the compact product space Zm. Therefore, for each ε>0 there exists a finite subset L in D such that for each x ∈ D there exists y ∈ L verifying d(gi(x), gi(y)) < ε for 1 ≤ i ≤ m. Step 2. Since S := {(a1,a2,...,an) : α = (am) ∈ Σ,n ∈ N} is countable, there exists a bijection Ψ : N → S. For each n ∈ N,setDn := CΨ(n). By Step 1 applied to f and D1, there exists a finite set L1 ⊂ D1 such that for each x ∈ D1 there exists y ∈ L1 for which d(f (x), f (y)) < 1. Since f is in the closure of K , there exists g1 ∈ K such that for each y ∈ L1 we have d(g1(y), f (y)) < 1. Applying Step 1 to Y the finite subset {f,g1} of Z and D2, we obtain a finite subset L2 ⊂ D2 such that −1 for each x ∈ D2 there exists y ∈ L2 with d(f (x), f (y)) < 2 ,d(g1(x), g1(y)) < −1 2 . Since f belongs to the closure of K , there exists g2 ∈ K such that for each −1 y ∈ L1 ∪ L2 we have d(g2(y), f (y)) < 2 . Step 1 applied to the subset {f,g1,g2} Y of Z and D3 provides a finite subset L3 ⊂ D3 such that for each x ∈ D3 there exists y ∈ L3 that verifies

~~−1 −1 −1 d(f (x), f (y)) < 3 ,d(g1(x), g1(y)) < 3 ,d(g2(x), g2(y)) < 3 . 112 4 Web-Compact Spaces and Angelic Theorems~~

Again, as f belongs to the closure of K , there exists g3 ∈ K such that for each −1 y ∈ L1 ∪ L2 ∪ L3 we have d(g3(y), f (y)) < 3 . We continue this procedure by induction. Consequently, we proved two facts: Fact 1. If (np)p is an increasing sequence of natural numbers such that x ∈ −1 Dnp , and if znp is the element of Lnp such that d(f(x),f(znp )) < np and −1 d(gi(x), gi(znp )) < np for i

~~f(x)= lim f(zn ), (4.1) p→∞ p and~~

~~gi(x) = lim gi(zn ) (4.2) p→∞ p for each i ∈ N. −1 ∈ ∪ ∪···∪ Fact 2. Since d(g n(y), f (y)) < n for each y L1 L2 Ln, we deduce ∈ that for each y i Li we have~~

~~f(y)= lim gn(y). (4.3) n→∞~~

~~Step 3. We prove that f(x)= limn→∞ gn(x) for each x ∈ Y . Since (Z, d) is a compact metric space, we only need to prove that if (nk)k is an increasing sequence of natural numbers and~~

w = lim gn (x), (4.4) k→∞ k then f(x)= w. Let α = (an)n ∈ Σ be such that x ∈ Aα. Then we may determine an increasing ∈ := sequence (np)p of natural numbers such that x Dnp Ca1a2...anp and, addition- := ally, we may assume that, if p

~~f(x)= lim f(zn ) = lim f(yp) = lim lim gn(yp). (4.5) p→∞ p p→∞ p→∞ n→∞~~

~~We rewrite (4.2) in the form~~

~~gi(x) = lim gi(yp), (4.6) p→∞ and from (4.6) and (4.4) we deduce~~

~~w = lim gn (x) = lim lim gn (yp). (4.7) k→∞ k k→∞ p→∞ k~~

~~Clearly, (4.5) implies~~

~~f(x)= lim lim gn (yp). (4.8) p→∞ k→∞ k 4.3 Web-compact spaces 113~~

Applying (4.7) and (4.8), and using the hypothesis concerning the interchangeable double limits, we note f(x)= w. This and the compactness of the metric space (Z, d) ensure f(x)= limn→∞ gn(x) for each x ∈ Y . The proof is completed.

~~4.3 Web-compact spaces~~

This section deals with a large class of topological spaces X for which Cp(X) is angelic. Most results presented here are from [320]. Following Orihuela [320], a topological space X is said to be web-compact if N there exists a nonempty subset Σ of N and a family {Aα : α ∈ Σ} of subsets of := { : X (called a web-compact representation for X) such that, if Cn1,n2,...,nk Aβ β = (mk) ∈ Σ,mj = nj ,j = 1, 2,...,k}, for every α = (nk) ∈ Σ, the following two conditions hold: (i) {Aα : α ∈ Σ}=X. = ∈ ∈ ∈ N (ii) If α (nk) Σ and xk Cn1,n2,...,nk for all k , then the sequence (xk)k has a cluster point in X. The following proposition characterizes web-compact spaces.

Proposition 4.1 For a topological space X, the following conditions are equivalent: (i) X is web-compact. (ii) There exist a metrizable and separable space P and a map T from P into P(X)such that; (a) {T(x): x ∈ P } is dense in X and (b) if (xn)n converges in P , then {T(xn) : n ∈ N} is a relatively countably compact set in X.

N Proof (i) ⇒ (ii): Set P := Σ endowed with the topology of N .LetT(α):= Aα. (ii) ⇒ (i): Since every Polish space is a continuous image of NN, there exists a N continuous map φ from a subset Σ ⊂ N onto P .Forα ∈ Σ,setAα := Tφ(α). Then {Aα : α ∈ Σ} is a web-compact representation in X.

~~It turns out that the class of web-compact spaces is reasonably rich. We provide some examples of web-compact spaces.~~

~~Example 4.1 The following spaces are web-compact.~~

N (1) Let {Aα : α ∈ N } be a family of relatively countably compact sets in a space X N whose union Y is dense in X.Let{Aα : α ∈ N } be a resolution in Y . It is easy to see N that X is web-compact. Indeed, if (αn)n is a sequence in N that converges to α ∈ NN ∈ NN ≤ ∈ N ⊂ , there exists β such that αn β for all n . Therefore n Aαn Aβ . Now Proposition 4.4 applies. Consequently, every space X admitting a sequence of 114 4 Web-Compact Spaces and Angelic Theorems relatively countably compact sets whose union is dense in X is web-compact. In particular, all separable spaces are web-compact. (2) Any Lindelöf Σ-space is web-compact. Hence K-analytic spaces are web- compact. (3) Quasi-Suslin spaces are web-compact. On the other hand, note that there are web-compact spaces that are not quasi- Suslin. Indeed, the space RR is web-compact (as it is separable), but RR is not quasi-Suslin by Proposition 3.8. Monterde Pérez [300, Proposition 80] showed that, if E is an lcs such that (E ,σ(E ,E)) is web-compact, then on E there exists a { : ∈ } web-compact representation Dα α Σ such that α∈Σ Dα is a linear subspace of E . = Floret [165, Theorem 3.7] proved that, if X n Kn, where the sets Kn are relatively countably compact, and Z is a metric space, the space Cp(X, Z) is angelic (see also Proposition 4.2 below). This result summarized some results obtained earlier by Grothendieck, Fremlin, Pryce and De Wilde (see [165] for references). The following result [320, Theorem 3] extends [165, Theorem 3.7 and Corollary 3.1].

~~Theorem 4.5 (Orihuela) For a web-compact space X, the space Cp(X) is angelic.~~

Proof Let (Z, d) be a metric compactiﬁcation of the space of the real numbers. Clearly, the inclusion map i : Cp(X) → Cp(X, Z) is continuous and injective. Let {Aα : α ∈ Σ} be a family of subsets of X giving on X a web-compact structure. Set Y Y := {Aα : α ∈ Σ}.LetZ be the topological product of Y -copies of the space Y Z. Clearly the restriction map θ : Cp(X, Z) → Z is also continuous. Since the injective map

Y Φ := θ · i : Cp(X) → Z is continuous, it is enough to show (by Corollary 4.1) that if A is a relatively countably compact subset of Cp(X), and f belongs to the closure of K := Φ(A), then there exists a sequence (gn)n in A such that f(y)= limn gn(y) for each y ∈ Y . Clearly, it is enough to prove that there exists a sequence (Φ(gn))n in K such that

f(y)= lim Φ(gn(y)) = lim gn(y) n n for each y ∈ Y. This follows from Theorem 4.4. Indeed, if α = (am)m ∈ Σ, ∈ (mp)p is an increasing sequence of natural numbers, and yp Ca1a2...amp for p ∈ N, then (yp)p has an adherent point x ∈ X. Therefore, if (fn)n is a sequence in A and the limits involved in limn limp Φ(fn(yp)) = limn limp fn(yp) and limp limn Φ(fn(yp)) = limp limn fn(yp) exist, then both limits are equal. Indeed, since A is relatively countably compact, it follows that, if f is an adherent point of the sequence (fn)n, the two preceding double limits are equal to f(x).

We also refer the reader to [83] for more general results concerning applications of Theorem 4.5, which can also be applied to get the following fact; see [320, Corol- lary 1.3]. 4.3 Web-compact spaces 115

~~Proposition 4.2 If X is web-compact and Y is a metric space, the space Cp(X, Y ) of continuous functions from X to Y with the pointwise topology is angelic.~~

~~Proof By Theorem 4.5, the space Cp(X) is angelic. On the other hand, applying Fremlin’s theorem (Theorem 4.3) the space Cp(X, Y ) is also angelic.~~

~~To show another consequence of Theorem 4.5, we need the following fact (due to Grothendieck); see [165, Theorem 4.2] and Corollary 4.13 below.~~

~~Proposition 4.3 Let X be a compact space. Let B be the closed unit ball in the Banach space C(X). Then every compact set in B for the topology of Cp(X) is compact in the weak topology of C(X).~~

Proof Since Cp(X) is angelic, it is enough to show that every sequence in B that converges in Cp(X) also converges in the weak topology of C(X). This follows from the Lebesque dominated convergence theorem.

~~The following useful result is due to Talagrand [388, Theorem 3.4].~~

~~Proposition 4.4 (Talagrand) Let X be a compact space. Then Cp(X) is K-analytic if and only if the weak topology of the Banach space C(X) is K-analytic.~~

~~Proof Let τp be the original topology of Cp(X). Since the weak topology of Cp(X) coincides with τp, the identity map~~

~~I : (C(X), σ (C(X), C(X) )) → Cp(X)~~

is continuous. Hence, if (C(X), σ (C(X), C(X) )) is K-analytic, then Cp(X) is K- analytic. N To prove the converse, assume Cp(X) is K-analytic. Let {Kα : α ∈ N } be a compact resolution in Cp(X). We prove that the closed unit ball B in the Ba- nach space C(X) is a K-analytic subspace of the space (C(X), σ (C(X), C(X))). N Clearly, {Kα ∩ B : α ∈ N } is a compact resolution in τp. Since Cp(X) is angelic by Theorem 4.5, the space B is angelic in τp. Hence, by Corollary 4.2, the space N (B, σ (C(X), C(X) )|B) is angelic. Note that {Kα ∩ B : α ∈ N } is a compact resolution in (B, σ (C(X), C(X) )|B) by Proposition 4.3. Now Corollary 3.6is ap- = plied to deduce that B is K-analytic in σ(C(X),C(X)). Since C(X) n Bn, where Bn := nB for each n ∈ N, and each set Bn is K-analytic in the topology σ(C(X),C(X)), the space (C(X), σ (C(X), C(X))) is K-analytic as claimed.

~~Another consequence of Theorem 4.5 is the following interesting fact; see [320].~~

~~Corollary 4.3 Let E be an lcs such that (E,σ(E,E)) is web-compact. Then the space (E, σ (E,E))is angelic. 116 4 Web-Compact Spaces and Angelic Theorems~~

~~Proof This follows from Theorem 4.5 and the fact that (E, σ (E, E)) is a subspace of Cp((E ,σ(E , E))).~~

Corollary 4.3 is applied to show that the weak topology σ(E,E) of a metrizable lcs is angelic. Indeed, (E,σ(E,E)), as covered by a sequence of compact sets, is web-compact. Note that the Mackey dual (E,μ(E,E))of a Banach space E need not be angelic. Indeed, if E is a Banach space and F ⊂ E is μ(E,E)-sequentially compact, the μ(E,E)-closure of F is μ(E,E)-precompact and complete; see [207], [367]. By [301], for a Fréchet space E,anyμ(E,E)-relatively countably compact set of E is μ(E ,E)-relatively compact, and for E := 1[0, 1] there exists a μ(E,E)-sequentially compact set in E that is μ(E,E)-relatively compact and not μ(E,E)-compact. Hence (E,μ(E,E))is not angelic. On the other hand, if E is a separable lcs, the space E admits a metric topology that is weaker than σ(E,E). Hence σ(E,E), and then μ(E,E), is angelic by Corollary 4.2.IfE is a WCG Banach space (which need not be separable), σ(E,E) is K-analytic (see Theorem 12.8 below). Consequently, (E ,σ(E ,E)) ⊂ CP (E, σ (E, E )) and angelic by Theorem 4.5. Hence (E,μ(E,E))is angelic as well. By T we denote the torus {z ∈ C :|z|=1}. For a topological Abelian group E, by E∧ we denote the set of all continuous group homomorphisms ϕ : E → T.The ∧ coarsest group topology on E for which all elements of E are continuous is called the Bohr topology and is denoted by σ E,E∧ .

~~Proposition 4.5 Let E be a metrizable dual-separating Abelian topological group. Then (E, σ (E, E∧)) is angelic.~~

Proof Let (Un)n∈N be a decreasing basis of neighborhoods of zero in E. Then := { ∈ ∧ : ⊂ T } Un φ E φ(Un) + , where T+ := {z ∈ C :|z|=1, Re z 0} is closed in the compact space TE and ∧ ∧ = hence compact in σ(E ,E), and each Un is equicontinuous. Note that E ∈ ∧ −1 T n Un . Indeed, if φ X , then φ ( +) is a neighborhood of zero, so there ∈ N ⊂ −1 T ∈ exists m such that Um φ ( +). Therefore φ Um . This proves that ∧ ∧ ∧ ∧ (E ,σ(E ,E)) is web-compact. Cp((E ,σ(E , E)), T) is angelic by Proposi- ∧ ∧ ∧ tion 4.2. Since (E, σ (E, E )) ⊂ Cp((E ,σ(E , E)), T), the proof is completed.

~~4.4 Subspaces of web-compact spaces~~

~~Proposition 4.6 from [83, Theorem 1] shows some relations between web-compact spaces and Lindelöf Σ-spaces.~~

Proposition 4.6 Let X be a web-compact regular space. Let {Aα : α ∈ Σ} be a web-compact representation of X such that every relatively countably compact set in 4.4 Subspaces of web-compact spaces 117

~~X is relatively compact. Then X contains a dense Lindelöf Σ-subspace. If Σ = NN, then X contains a dense K-analytic subspace.~~

Proof For α = (an) ∈ Σ,set := { ∈ :∃ ⊂ ∈ ∈ { : ∈ N}} Bα x X (xn)n X, xn Ca1,...,an ,x xn n . ⊂ ∈ := Clearly, Aα Bα for each α Σ, and Y α∈Σ Bα is a dense subspace of X.The reader will check that each set Bα is relatively countably compact in X; therefore Bα is relatively compact (by the assumption). We prove that Y is a Lindelöf Σ-space by showing that the map T from Σ into compact subsets of Y , deﬁned by T(α):= Bα, is usco. Let U be a closed (by regularity of X) neighborhood of T(α). There exists ∈ N ⊂ m such that Ca1,...,am U. Indeed, otherwise there exists a sequence ∈ \ xn Ca1,...,an U for each n ∈ N. Since {Aα : α ∈ Σ} is a web-representation of X, the sequence (xn)n has an adherent point x ∈ Bα, a contradiction. This proves that, for each β = (bn) ∈ Σ such that aj = bj for j = 1,...,m,wehave = ⊂ ⊂ ⊂ Tβ Bβ Cb1,...,bm Ca1,...,am U, and consequently T is usco.

A compact space X is called Talagrand compact if C(X) is K-analytic in its weak topology [147]. By Proposition 4.4, a compact space X is Talagrand compact if and only if Cp(X) is K-analytic. Proposition 4.6 is applied to prove a couple of consequences observed in [82] and presented below.

~~Corollary 4.4 Let X be a web-compact space with a web-compact representation N for the case Σ = N . Then every compact set K ⊂ Cp(X) is a Talagrand compact space.~~

Proof The map δ : X → Cp(K) deﬁned by δ(x) := δx , where δx(f ) = f(x)for all f ∈ K, is continuous. This shows that δ(X) ⊂ Cp(K) is a web-compact angelic and regular space. Proposition 4.6 provides in δ(X) a dense K-analytic subspace Y that separates points of K. By Theorem 14.1, the space Cp(K) is K-analytic, so K is Talagrand compact.

Corollary 4.5 Let X be a web-compact space. Assume that K ⊂ Cp(X) is a compact set. The following assertions are equivalent: (i) K is metrizable. (ii) K is contained in a separable subset of Cp(X).

Proof (i) ⇒ (ii) is clear. (ii) ⇒ (i): Let D be a countable subset of Cp(X) such that K ⊂ D (the closure in Cp(X)). Deﬁne the map δ : X → Cp(D) by the formula δ(x) := δx , where 118 4 Web-Compact Spaces and Angelic Theorems

δx(f ) := f(x) for x ∈ X and f ∈ D. Clearly, δ is continuous, δ(X) is a web- compact subspace of the angelic (Theorem 4.5) regular space Cp(D), and Cp(D) admits a weaker metrizable topology ξ (since D is separable). Note also that δ(X) is separable. Indeed, by Proposition 4.6 the space δ(X) contains a dense Lindelöf Σ-subspace Y ,soY is Lindelöf. Therefore (Y, ξ|Y) is separable and metrizable. Now Lemma 3.1 is applied to derive that Y in its original topology is separable. Hence δ(X) is separable, too. Finally, observe that K can be considered a sub- ∧ set of Cp(δ(X)). Indeed, for every f ∈ K, deﬁne the function f on δ(X) by ∧ ∧ ∧ ∧ ∧ f (δx) := f(x), and set K := {f : f ∈ K}. Clearly, K ⊂ Cp(δ(X)), and K is homeomorphic to K. Since Cp(δ(X)) admits a weaker metrizable topology, we reach the conclusion.

Corollary 4.6 Let E be an lcs such that (E,σ(E,E)) contains a dense web- compact subset X. Then (E, σ (E, E)) is angelic, and every weakly compact subset of E is metrizable if and only if it is contained in a separable subspace of E. For the case Σ = NN, every weakly compact subset of E is Talagrand compact.

Proof By Theorem 4.5, the space Cp(X) is angelic. Let A be a relatively countably compact subset of (E, σ (E, E )), and let Φ : (E, σ (E, E )) → Cp(X) be a continuous and injective map deﬁned by the restriction. Then Φ(A) is a relatively countably compact subset of Cp(X). By the angelicity of Cp(X),wehavethatΦ(A) is compact, and for each subset B of Φ(A) and each b ∈ B there exists a sequence (bn)n in B such that b = limn bn. By Theorem 4.1,thesetΦ(A) is closed because Φ(A) = Φ(A) and Φ|A is a homeomorphism. Therefore A is a compact subset of (E, σ (E, E )). Moreover, for each a ∈ A there exists a sequence (an)n in A such N that a = limn an. This proves that (E, σ (E, E )) is angelic. If Σ = N , by Corol- lary 4.4 every weakly compact set in E is Talagrand compact, and Corollary 4.5 completes the proof.

Note that the class of lcs’s E whose weak∗ dual (E,σ(E,E))includes a dense web-compact subset contains a large class G of lcs’s (see Chapter 11). Therefore, in this particular case, the conclusions of Corollary 4.6 hold for spaces in the class G (which includes all (DF )- and (LM)-spaces).

~~4.5 Angelic duals of spaces C(X)~~

This short section provides another application of Corollary 4.6 to study the angelicity of the dual of spaces C(X).LetCb(X) be the space of all continuous bounded real-valued functions on X.Bythestrict topology β0 on Cb(X),wemeanalocally convex topology generated by the family of seminorms (ph)h deﬁned by

ph(f ) := sup |f(x)h(x)|, x∈X 4.5 Angelic duals of spaces C(X) 119 where f ∈ Cb(X) and h runs over the space of real-valued bounded functions on X vanishing at inﬁnity (i.e., for each ε>0 there exists a compact set K ⊂ X such that |h(x)| <εfor each x ∈ X \ K;see[70]). It is known (see [70]) that the dual (Cb(X), β0) = Mt (X) is the space of all ﬁnite Borel measures on X that are inner regular with respect to the compact sets. We need the following simple lemma from [83, Lemma 13].

~~Lemma 4.1 For a uniform space (X, U), the space Ub(X) is a dense subspace of (Cb(X), β0).~~

Proof Recall that Ub(X) := {f ∈ Cb(X) : f is U − uniformly continuous}.Itis ∈ = ∈ = enough to prove that, if μ Mt (X) and X gdμ 0 for all g Ub(X), then μ 0. Fix f ∈ Cb(X) and ε>0. Let |μ| be the variation of μ. Then there exists compact K ⊂ X such that |μ|(X \K) < ε. Since the restriction f |K is uniformly continuous, we apply [146, Theorem 8.5.6] to get a function g ∈ Ub(X) such that f |K = g|K = := | | = and g ∞ f ∞, where f ∞ supx∈X f(x). Then X gdμ 0 and fdμ = (f − g)dμ ≤ |f − g|dμ ≤ 2f ∞ε. X X X\K

N If a uniform space (X, U) admits a U-basis (i.e., a basis {Nα : α ∈ N } such that Nα ⊂ Nβ for β ≤ α), then precompact subsets of (X, U) are metrizable. This result will be proved in Proposition 6.11 below by using the concept of trans-separability; see also [82, Theorem 1]. Proposition 6.11 will be used for the proof of Proposi- tion 4.7, due to Cascales and Orihuela [83, Theorem 14].

~~Proposition 4.7 Let (X, U) be a uniform space with a U-basis. Then the dual Mt (X) is σ(Mt (X), Cb(X))-angelic, and each σ(Mt (X), Cb(X))-compact set in Mt (X) is metrizable.~~

~~N Proof For α = (an) ∈ N ,set ={ ∈ : ≤ | − |≤ −1 ∈ ∈ N} Aα f Cb(X) f ∞ a1, f(x) f(y) n ,(x,y) N(an.....),n .~~

Since the compact-open topology τc of Cb(X) coincides with β0 on uniformly bounded subsets of Cb(X), and each Aα uniformly bounded and uniformly equicontinuous is compact in τc (by Ascoli’s theorem [146]), we deduce that Aα is β0- N compact. Note that {Aα : α ∈ N } is a compact resolution in Ub(X); this will show (by applying Lemma 4.1) that (Cb(X), β0) is web-compact. Indeed, choose arbi- ∈ = k ∈ N NN trary f Ub(X). Then there exist L>0 and a sequence αk (an), k ,in such that ≤ | − |≤ −1 ∈ f ∞ L, f(x) f(y) k , (x,y) Nαk . N Deﬁne α = (an) ∈ N by the formula := { 1} := { 1 2 n} a1 max L,a1 ,an max an,an−1,...,a1 , 120 4 Web-Compact Spaces and Angelic Theorems for all n ≥ 2. Then f ∈ Aα. The ordering condition required for a family to be a resolution is obvious. Consequently, using Corollary 4.6, we note that

(Mt (X), σ (Mt (X), Cb(X))) is angelic. Note that every compact set K in (Mt (X), σ (Mt (X), Cb(X))) is metrizable. Indeed, if K is compact in the topology τ on Mt (X) of the uniform convergence on the sets Aα, by Proposition 6.11 (already mentioned above) K is τ -metrizable. If τ0 is the topology on Mt (X) of the uniform convergence on uniformly bounded and equicontinuous sets of Cb(X),wehaveτ ≤ τ0, and σ(Mt (X), Cb(X)) ≤ τ0. Observe that K is also τ0-compact. Indeed, since τ0 is angelic (see Theorem 4.1), it is enough to show that K is τ0-sequentially compact: If ∈ (μn)n is a sequence in K, there is μ K and a subsequence (μnk )k converging to μ → in σ(Mt (X), Cb(X)). Then, by [143, Theorem 7], we have μnk μ in τ0. Hence σ(Mt (X), Cb(X))|K = τ|K = τ0|K, and the proof is completed.

~~4.6 About compactness via distances to function spaces C(K)~~

In this section, which supplements the previous one, we survey some classical results about compactness concepts by using suitable inequalities involving distances to spaces of continuous functions. We present some quantitative versions of Grothendieck’s characterization of the weak compactness for spaces C(K) for compact Hausdorff spaces K and quantitative versions of the classical Eberlein– Grothendieck and Krein–Šmulian theorems. The ﬁrst part of this section will be used to present corresponding results for general Banach spaces E. Results of this section are based on recent works of Angosto and Cascales [6], [7], [10], Angosto [9], Angosto, Cascales and Namioka [8] and Cascales, Marciszewski and Raja [92]. In the case of Banach spaces, these quantitative generalizations have been previously studied by Fabian, Hájek, Montesinos and Zizler [150] and Granero [187]. Let (Z, d) be a metric space. Let A be a nonempty subset of Z. Set

~~diam(A) := sup{d(x,y) : x,y ∈ A}.~~

By the distance between nonempty sets A,B in Z we mean d(A,B) := inf{d(a,b) : a ∈ A,b ∈ B}. By the Hausdorff nonsymmetrized distance from A and B, we mean d(A,B)ˆ := sup{d(a,B) : a ∈ A}. The product space ZX for a space X will be considered with the standard supremum metric

d(f,g) := sup d(f (x), g(x)), x∈X which is allowed to take the value +∞. For a subset A of a space X,byclustX(φ) N we mean the set of all cluster points in X of a sequence φ ∈ A . Recall also that = { : } clustX(φ) n φ(m) m>n . 4.6 About compactness via distances to function spaces C(K) 121

For a compact space K and ε ≥ 0, we will say that a set H ⊂ C(K) ε- interchanges limits with K (see [150]) if lim lim fn(xk) − lim lim fn(xk) ≤ ε n k k n for any two sequences (xn)n in K and (fm)m in H , provided the iterated limits exist. This concept for ε = 0 is due to Grothendieck; see [165]. Let K be a compact Hausdorff space. Let τp be the pointwise topology on the space RK , and let d be the metric of the uniform convergence on RK .LetH ⊂ RK be a pointwise bounded set. By the classical Tichonov theorem, the closure H of H in τp is τp-compact. Hence, to show that H is τp-relatively compact in C(K), it is enough to see when H ⊂ C(K). Assume that dˆ := d(ˆ H,C(K))is the nonsymmetrized distance from H ˆ to C(K). Then d>0 provides us a non-τp-compactness measure for H relative to the space C(K). Clearly, H ⊂ C(K) ⇔ dˆ = 0.

A pointwise bounded set H is a relatively compact set in Cp(K) if and only if d(ˆ H,C(K))= 0. It is important to know how to compute dˆ. The distance of a function f ∈ RK to the space C(K) can be obtained by the following easy formula in Theorem 4.6; see [10]. The proof of Theorem 4.6 (from [10, Theorem 2.2]) uses the following fact [212, Theorem 12.16].

Lemma 4.2 Let X be a normal space, and let f1 ≤ f2 be two real functions on X such that f1 is upper semicontinuous and f2 is lower semicontinuous. Then there exists on X a continuous function f such that f1 ≤ f ≤ f2.

~~Theorem 4.6 Let X be a normal space and let f ∈ RX. Then − d(f,C(X)) = 2 1 osc(f ), := = { : ⊂ ∈ } where osc(f ) supx∈X osc (f, x) supx∈X inf diam f(U) U X open, x U .~~

Proof First we prove − 2 1 osc(f ) ≤ d(f, C(X)). It is enough to check this inequality for ﬁnite d := d(f,C(X)).Fixx ∈ X and ε>0. Then there exists g ∈ C(X) such that − d(f,g) ≤ d + 3 1ε.

~~By the continuity of g, there exists an open neighborhood U of x such that diam(g(U)) < 3−1ε. Then~~

d(f (y), f (z)) ≤ d(f (y), g(y)) + d(g(y), g(z)) + d(g(z), f (z)) < 2d + ε 122 4 Web-Compact Spaces and Angelic Theorems for all y,z ∈ U. Hence osc(f, x) ≤ 2d for each x ∈ X, and the claim holds. Next, we prove − d(f,C(X)) ≤ 2 1 osc(f ). −1 It is enough to show this for ﬁnite t := 2 osc(f ).Forx ∈ X,letUx be the family of open neighborhoods of x. Set

~~Vx := {U ∈ Ux : diam (f (U)) < osc(f ) + 1}.~~

~~Note that the restricted map f |U is upper and lower bounded for each U ∈ Vx , and Vx forms a basis of neighborhoods of x in X. Then~~

~~inf sup f(y)− sup inf f(z)= ∈ z∈U U Vx y∈U U∈Vx~~

~~inf sup (f (y) − f(z))≤ U,V∈Vx y∈U,z∈V~~

~~inf sup (f (y) − f(z))= inf diam(f (U)) = U∈Vx y,z∈U U∈Vx~~

~~inf diam(f (U)) = osc(f, x) ≤ 2t. U∈Ux Now set~~

~~f1(x) := inf sup f(z)− t, U∈Vx z∈U~~

~~f2(x) := sup inf f(z)+ t. z∈U U∈Vx~~

~~Then f1 ≤ f2, f1 is upper semicontinuous and f2 is lower semicontinuous. Now Lemma 4.2 is applied to get a continuous function h such that f1 ≤ h ≤ f2 on X. Since~~

~~f2(x) − t ≤ f(x)≤ f1(x) + t for each x ∈ X,wehave~~

~~h(x) − t ≤ f2(x) − t ≤ f(x)≤ f1(x) + t ≤ h(x) + t.~~

~~Thus d(f,h) ≤ t = 2−1 osc(f ).~~

Corollary 4.7 Let X be a topological space. The following conditions are equivalent: (i) X is a normal space. (ii) For each f ∈ RX, there exists g ∈ C(X) such that d(f,g) = 2−1 osc(f ). (iii) d(f,C(X)) = 2−1 osc(f ) for each f ∈ RX. 4.6 About compactness via distances to function spaces C(K) 123

~~Let X be a topological space and H ⊂ RX. Deﬁne~~

~~ck(H ) := sup d(clustRX (φ), C(X)). φ∈H N~~

For K ⊂ X, deﬁne γK (H ) as sup d lim lim fm(xn), lim lim fm(xn) : (fm)m ⊂ H,(xn)n ⊂ K n m m n provided the iterated limits exist. If H ⊂ C(X) is τp-relatively countably compact in C(X), then ck(H ) = 0. The equality γK (H ) = 0 means that in H we interchange limits with K. The following formula ([6, Theorem 2.3]; see also [92], [10]) describes the distance dˆ := d(ˆ H,C(K)) (the closure in RK ) by using the quantities above. We present only a sketch of the proof.

~~Theorem 4.7 (Angosto–Cascales) Let K be a compact space. Let H ⊂ C(K) be a uniformly bounded set. Then ˆ ck(H ) ≤ d(H,C(K))≤ γK (H ) ≤ 2ck(H ), where the closure is taken in RK .~~

Proof The ﬁrst inequality, ck(H ) ≤ d(ˆ H,C(K)), follows from the deﬁnitions. To prove the inequality γK (H ) ≤ 2ck(H ), we use a standard argument to show that τp-relative compactness of H in C(K) implies that H interchanges limits with K. Indeed, let (fm)m be a sequence in H ,let(xn)n be a sequence in K, and assume that both iterated limits

lim lim fm(xn), lim lim fm(xn), n m m n exist in R.Fixanyt ∈ R such that ck(H ) < t. Then the sequence (fm)m has K a τp-cluster point f ∈ R such that d(f,C(K)) < t.Fixg ∈ C(K) such that | − | ∈ supx∈K f(x) g(x)

~~lim g(xn ) = g(x), k k and fm(x) is the limit of the convergent sequence (fm(xnk ))k. Hence lim f(xn ) − f(x) ≤ lim f(xn ) − lim g(xn ) +|g(x) − f(x)|≤2t k k k k k k and~~

~~lim lim fm(xn) = lim fm(x) = f(x). m n m 124 4 Web-Compact Spaces and Angelic Theorems~~

Consequently, lim lim fm(xn) − lim lim fm(xn) = lim lim fm(xn) − f(x) = n m m n n m lim f(xn ) − f(x) ≤ 2t. k k ˆ Now we prove that d(H,C(K))≤ γK (H ). Deﬁne ∗ osc (f, x) := inf sup |f(y)− f(x)|:U ⊂ X open ,x ∈ U U y∈U

~~∗ for each x ∈ X and f ∈ H . Then osc (f, x) ≤ γK (H ) and osc(f ) ≤ 2γK (H ). ˆ We apply Theorem 4.6 to deduce d(f,C(K)) ≤ γK (H ). Finally, d(H,c(K)) ≤ γK (H ).~~

~~We note the following Eberlein–Grothendieck result (see [165]) as a consequence of the results above.~~

Corollary 4.8 Let K be a compact space and H ⊂ C(K). The following assertions are equivalent: (i) H is τp-relatively countably compact in C(K). (ii) H interchanges limits with K. (iii) H ⊂ C(K), where the closure is taken in RK . (iv) H is τp-relatively compact in C(K).

The next result, due to Cascales, Marciszewski and Raja, [92, Theorem 3.3], says that the ε-interchanging limit property is preserved when taking convex hulls. We refer also to [150, Theorem 13], where a similar result has been proved for subsets in Banach spaces. The proof of Theorem 4.8, originally presented in [92], uses some ideas from the proof of the Krein–Šmulian theorem; see [240, Chapter 5]. We need the following two lemmas due to Kelley and Namioka [240, Lemma 17.9, Lemma 17.10]. By P(X) we denote the power set of a set X. The proof of Lemma 4.3 is omitted, and we refer the reader to the book [240].

Lemma 4.3 Let μ be a ﬁnitely additive ﬁnite measure on an algebra of sets A , and let (An)n be a sequence of sets in A such that μ(An)>δfor some δ>0 and ∈ N m all n . Then there exists a subsequence (Ank )k such that μ( i=1 Ani )>0 for each m ∈ N.

Lemma 4.4 Let (In)n be a sequence of pairwise disjoint ﬁnite nonempty sets, and let μn be a probability measure on P(In) for each n ∈ N. Let (Ak)k be a sequence := ∩ of subsets of I n In such that for some δ>0 we have lim infn μn(Ak In)>δ ∈ N j = ∅ for each k . Then there exists a subsequence (Aki )i such that i=1 Aki for each j ∈ N. 4.6 About compactness via distances to function spaces C(K) 125

Proof We denote by A the (countable) subalgebra of P(I) generated by the sets N ∩ An. Then there exists an increasing sequence (nk)k in such that limk μnk (A Ink ) exists for each A ∈ A . Deﬁne a ﬁnitely additive measure on A by the formula

~~μ(A) := lim μn (A ∩ In ) k k k for A ∈ A . Since μ(Ak)>δfor each k ∈ N, we apply Lemma 4.3 to complete the proof.~~

~~We are ready to prove the following theorem.~~

Theorem 4.8 (Cascales–Marciszewski–Raja) Let Z be a compact convex subset of a normed space E. Let K beaset, and let H be a subspace of ZK . Then, for each ε ≥ 0, the space Hε-interchanges limits with K if and only if conv(H ) ε- interchanges limits with K.

~~Proof Let (fn)n and (xk)k be sequences in conv(H ) and K, respectively, with the property that both limits~~

~~lim lim fn(xk), lim lim fn(xk), n k k n exist. Set γ := lim lim fn(xk) − lim lim fn(xk). (4.9) n k k n~~

For each n ∈ N, there exist gs ∈ H and ts ∈[0, 1] such that fn = tsgs, ts = 1. s∈In s∈In := We may assume that the sets In are pairwise disjoint. Set I n In.By(xk)k we denote again a subsequence of (xk)k such that, for every s ∈ I , gs(xk) → rs(∈ Z) for some r . Then p := lim f (x ) = t r . There exists ξ ∈ B such that s n k n k s∈In s s E

~~γ = ξ lim lim fn(xk) − lim lim fn(xk) = (4.10) n k k n~~

ξ lim pn − lim lim fn(xk) = lim ξ lim pn − lim fn(xk) , (4.11) n k n k n n where BE is the closed unit ball in the dual E of E. Fix arbitrary δ>0. Without loss of generality we may assume that for each k ∈ N

~~ξ lim pn − lim fn(xk) = lim ξ(pn − fn(xk)) > γ − δ. (4.12) n n n~~

~~Therefore, for each k ∈ N, there exists nk ∈ N such that for all n ≥ nk we have ξ(pn − fn(xk)) > γ − δ. For each k ∈ N,set~~

~~Ak := {s ∈ I : ξ(rs − gs(xk)) > γ − 2δ}. (4.13) 126 4 Web-Compact Spaces and Angelic Theorems~~

∈ N := Next, for each n , deﬁne a probability measure on In by the formula μn(A) ⊂ s∈A ts for each A In.LetM be the diameter of Z, and we may assume that M>0. Then γ − δ<ξ(pn − fn(xk)) = ξ tsrs − tsgs(xk) (4.14) s∈In s∈In = tsξ(rs − gs(xk)) (4.15) s∈In = tsξ(rs − gs(xk)) + tsξ(rs − gs(xk)) ≤ (4.16) s∈In∩Ak s∈In\Ak tsM + γ − 2δ = μn(In ∩ Ak)M + γ − 2δ (4.17) s∈In∩Ak −1 for each k ∈ N and each n ≥ nk. This yields the inequality μn(In ∩ Ak)>δM . Hence −1 lim inf μn(In ∩ Ak) ≥ δM n

∈ N j = for all k . Now Lemma 4.4 applies to get a subsequence (Aki )i with i=1 Aki ∅ for all j ∈ N. Then, if j ∈ N, we can select sj ∈ I with − − ξ(rsj gsj (xki )) > γ 2δ for all i ≤ j. Set := := := hj gsj ,dj rsj ,yi xki , for i, j ∈ N. Then ξ(dj − hj (yi)) > γ − 2δ for all i, j ∈ N. We proceed as before with the sequence (hj )j . There exists a subsequence of (hj )j , which we denote again by (hj )j , such that limj hj (yi) = wi in Z for each i ∈ N. Also, the corresponding sequence (dj )j converges to some d ∈ Z. We select a subsequence of (yi)i , which we denote again by (yi)i , such that the corresponding sequence (wi)i converges to w ∈ Z. This yields

~~lim lim hj (yi) = lim dj = d, lim lim hj (yi) = lim wi = w, j i j i j i and it follows that~~

~~ξ(d − wi) ≥ γ − 2δ. We conclude that~~

~~lim ξ(d − wi) = ξ(d − w) ≥ γ − 2δ. i 4.6 About compactness via distances to function spaces C(K) 127~~

~~As ξ ∈ BE was arbitrary, d − w=lim lim hj (yi) − lim lim hj (yi) ≥ γ − 2δ. (4.18) j i i j~~

~~Since, by the assumption, the set Hε-interchanges limits with K, we deduce that ε ≥ γ − 2δ. Hence ε ≥ γ since δ was arbitrary.~~

~~Corollary 4.9 For a compact space K and a uniformly bounded set H ⊂ C(K), we have d(ˆ conv(H ), C(K)) ≤ 2d(ˆ H , C(K)), where the closure is taken in RK .~~

~~Proof This follows from Theorem 4.8 and Theorem 4.7.Wehave ˆ ˆ d(conv(H ), C(K)) ≤ γK (conv(H )) = γK (H ) ≤ 2ck(H ) ≤ 2d(H , C(K)).~~

~~The following quantitative version of the classical Krein–Šmulian theorem follows also from the results above.~~

~~Theorem 4.9 (Cascales–Marciszewski–Raja) If K is a compact space and H ⊂ RK is uniformly bounded,~~

~~d(ˆ conv(H ), C(K)) ≤ 5d(ˆ H , C(K)), where the closure is taken in RK .~~

~~Proof Without loss of generality, we may assume that H is uniformly bounded and compact in RK since conv(H)= (conv(H )). Hence, we may assume that H = H . Therefore, it is enough to show that~~

~~d(ˆ conv(H ), C(K)) ≤ 5d(H,ˆ C(K)).~~

Take arbitrary ε>d(H,C(K))ˆ . Then, for each f ∈ H , one has d(f,C(K)) < ε, and then there exists gf ∈ C(K) such that f − gf ∞ ≤ ε. The set H0 := {gf : f ∈ C(H)} 4ε-interchanges with K. Indeed, note that H0 ⊂ H + B(0,ε), where the ∞ closed ball B(0,ε) is taken in the Banach space (K). Then H0 ⊂ H + B(0,ε). Since H ⊂ H0 + B(0,ε),we note (by using the facts above) that

~~H0 ⊂ H0 + B(0, 2ε). ˆ Consequently, d(H0, C(K)) ≤ 2ε. By Theorem 4.7, we conclude ck(H0) ≤ 2ε. Hence, again Theorem 4.7 is applied to get~~

~~γK (H0) ≤ 4ε. 128 4 Web-Compact Spaces and Angelic Theorems~~

~~This proves the claim. By Theorem 4.8, we know that~~

~~γK (conv(H0)) ≤ 4ε.~~

~~Again by Theorem 4.7,~~

~~conv(H0) ⊂ C(K) + B(0, 4ε).~~

~~Finally,~~

~~conv(H ) ⊂ conv(H0) + B(0,ε)⊂ C(K) + B(0, 5ε). The proof is completed.~~

~~Theorem 4.10 below provides a quantitative approach to show the already known fact stating that Cp(K) is angelic for a compact space K. We need the following useful lemma [92, Lemma 5.1].~~

Lemma 4.5 Let (Z, d) be a compact metric space, and let K be a set. Furthermore, K let f1,f2,...,fn ∈ Z and δ>0. Then there exists a ﬁnite subset L ⊂ K such that for each x ∈ K there exists y ∈ L such that d(fk(y), fk(x)) < δ for each 1 ≤ k ≤ n.

~~:= ∈ n Proof Set d∞((xk), (yk)) sup1≤k≤n d(xk,yk) for each (xk), (yk) Z . This de- ﬁnes a metric on compact Zn. Set~~

~~B := {f1(x), f2(x),...,fn(x) : x ∈ K}.~~

~~Note that (B, d∞) is totally bounded in (Zn,d∞). Therefore, there exists a ﬁnite set L ⊂ K such that {f1(x), f2(x), . . . , fn(x) : x ∈ L} is δ-dense in (B, d∞).~~

~~Now we are ready to prove the following interesting approximation theorem from [92, Proposition 5.2].~~

Theorem 4.10 (Cascales–Marciszewski–Raja) Let (Z, d) be a compact metric space, and let K be a set. Let H be a set in ZK that ε-interchanges limits with K. K Then, for each f ∈ H (the closure in R ) there exists a sequence (fn)n ⊂ H such that we have sup d(g(x), f (x)) ≤ ε x∈K K for any cluster point g of (fn)n in Z .

~~Proof For f , the corresponding sequence (fn)n will be defined by induction. Let ε ≥ 0 be fixed, and set f1 := f . By Lemma 4.5, there exists a finite set L1 ⊂ K such that~~

~~min d(f1(x), f1(y)) < 1 (4.19) y∈L1 4.6 About compactness via distances to function spaces C(K) 129~~

−1 for each x ∈ K. Since f ∈ H , there exists f2 ∈ H such that d(f2(y), f1(y)) < 2 for all y ∈ L1. Assume we have already deﬁned in H functions f1,f2,...,fn and the corresponding ﬁnite sets L1,L2,...,Ln (according to Lemma 4.5)forn ≥ 2 such that −1 min max {d(fk(x), fk(y))}

~~−1 max {d(fk(x), fk(yn))}~~

~~Then limn fk(yn) = fk(x) for ﬁxed x ∈ K and all k ∈ N. K Let g be a cluster point of (fk)n in Z .Let(fkj )j be a subsequence of (fk)k ∈ = such that for x K we have limj fkj (x) g(x). Also~~

~~lim lim fk (yn) = lim fk (x) = g(x) j n j j j and~~

~~lim lim fk (yn) = lim f1(yn) = f1(x) = f(x). n j j n Since H is a set that ε-interchanges limits with K, we get d(g(x), f (x)) ≤ ε. The proof is completed.~~

~~Theorem 4.10 can be used to prove the following corollary; see [240, Theo- rem 8.20] and [165, p. 31].~~

Corollary 4.10 Let Z be a compact metric space and let K beaset. Let H ⊂ ZK be a set that ε-interchanges limits with K for ε = 0. Then, for each f ∈ H there exists a sequence (fn)n in H such that fn(x) → f(x)for each x ∈ K.

~~Using Corollary 4.8 and Corollary 4.10, we obtain the following classical result; see [165] and compare Theorem 4.5.~~

~~Corollary 4.11 Cp(K) is angelic for any compact space K.~~

We complete this section with some applications for studying measures of weak noncompactness in Banach spaces; see [7]. We need the following concepts. For a bounded set H ⊂ E in a Banach space E, we deﬁne

~~ω(H) := inf {ε>0 : H ⊂ Kε + εBE,Kε ⊂ E is σ(E,E ) − compact}. 130 4 Web-Compact Spaces and Angelic Theorems~~

The function ω(H) was deﬁned by De Blasi [60] as a measure of weak noncompactness; see also [36], [41], [247], [248] for more information. By γ(H)we mean sup lim lim fm(xn) − lim lim fm(xn) : (fm)m ⊂ BE ,(xn)n ⊂ H , n m m n if the involved limits exist; see also [36], [247], [92], [150] for several situations where this concept has been used. Deﬁne

∗∗ k(H ) := d(ˆ H,E)= sup d(x ,E), x∗∗∈H where the closure is taken in the ω∗-topology of the bidual E (i.e., σ(E,E)), and d is the inf-distance for sets generated by the natural norm in E (we also refer the reader to works [92], [150], [187]). Finally, set

~~ck(H ) := sup d(clust E (φ), E), φ∈H N N ∗ where clust E (φ) denotes the set of all cluster points of φ ∈ H in (E ,ω ).Ob- serve that~~

k(H ) = inf {ε>0 : H ⊂ E + εBE }. (4.20) Note that k(H ) = 0 if and only if H is relatively weakly compact in E (since H in E is ω∗-compact, and then k(H ) = d(ˆ H,E)= 0 if and only if H ⊂ E if and only if H is weakly relatively compact). We need the following fact from [7, Proposition 2.1].

~~Lemma 4.6 Let H be a bounded set in a Banach space E. Then H 2ck(H )- interchanges limits with the dual ball BE in E .~~

Proof Fix a sequence (fm)m in BE .Let(xn)n be a sequence in H , and assume ∗ that both iterated limits exist in R.Fixt>ck(H ). Then (xn)n has a ω -cluster point z ∈ E such that d(z,E) < t.Letz ∈ E be such that z − z

Hence lim lim fm(xn) − lim lim fm(xn) = m n n m lim lim fm(xn) − f(z) = lim fm (z) − f(z) ≤ 2t. m n k k Theorem 4.11, originally from [92, Corollary 4.2], shows a way to transfer previous results to the context of Banach spaces. Its proof uses an argument from the proof of Theorem 4.6 and, instead of Lemma 4.2, the following fact from [98, The- orem 21.20]; see [92, Corollary 4.2].

~~Proposition 4.8 If f1~~

~~In order to prove Theorem 4.11, we need the following lemma; see [92].~~

Lemma 4.7 Let K be a convex compact set in an lcs E. Let A (K) be the set of all afﬁne real functions on K. Then d(f,C(K)) = d(f,A c(K)) for every bounded f ∈ A (K), where A c(K) := C(K) ∩ A (K).

~~Proof Note that d(f,C(K)) = 2−1 osc(f ) by Theorem 4.6. Thus it is enough to show that d(f,A c(K)) ≤ 2−1 osc(f ). Fix arbitrary ε>2−1 osc f , and set~~

f1(x) := inf sup{f(z): z ∈ U}−ε, f2(x) := sup inf{f(z): z ∈ U}+ε, U U where the inﬁmum in the ﬁrst equality and the supremum in the other one are taken over neighborhoods U of the point x. We prove that f1 is concave upper semicontinuous and f2 is concave lower semicontinuous. Indeed, we prove only the claim for the concavity of the function f1. The other case is similar. Let t>0, x,y ∈ K, and s ∈ (0, 1). Fix a neighborhood U of sx + (1 − s)y such that

~~sup {f(z): z ∈ U}−ε ≤ f1(sx + (1 − s)y) + t.~~

~~Next, choose neighborhoods V and W of x and y, respectively, such that sV + (1 − s)W ⊂ U. Then~~

~~sf1(x) + (1 − s)f1(y) ≤~~

~~s sup{f(z): z ∈ V }+(1 − s)sup{f(z): z ∈ W}−ε~~

~~≤ sup{f(z): z ∈ U}−ε ≤ f1(sx + (1 − s)y) + t. This proves the claim. 132 4 Web-Compact Spaces and Angelic Theorems~~

~~c Note that f1~~

~~sup |f(x)− h(x)|≤ε. x∈K Hence d(f,A c(K)) ≤ ε.~~

~~We are ready to prove Theorem 4.11.~~

Theorem 4.11 (Cascales–Marciszewski–Raja) Let E be a Banach space, and let BE be the closed unit ball in E endowed with the topology σ(E ,E). Let i : ∗∗ E → E and j : E → ∞(BE ) be the natural embeddings. Then d(x , i(E)) = ∗∗ ∗∗ d(j(x ), C(BE )) for each x ∈ E .

∗∗ Proof Since x ∈ E is afﬁne and bounded on BE , we apply Lemma 4.7 to get c ∗∗ ∗∗ a function h1 ∈ A (BE ) such that x − h1≤ε for ε>d(x ,C(BE )). Then ∗∗ ∗ ∗ x − h2≤ε, where h2(x ) =−h1(−x ) on BE . The function g : BE → R −1 c deﬁned by g := 2 (h1 + h2) belongs to A (BE ) and g(0) = 0. Moreover, ∗∗ −1 ∗∗ ∗∗ x − g≤2 x − h1+x − h2 ≤ ε. ∗∗ ∗∗ There exists a linear functional y on E such that y |BE = g. Then, by Grothendieck’s completeness theorem (see [213, Theorem 9.2.2]), there exists x ∈ E such that y∗∗ = i(x).Asx∗∗ − i(x)≤ε, the proof is completed.

∞ B Since (BE ) can be considered as a subspace of (R E ,τp), the map j : ∗ ∞ (E ,ω ) → ( (BE ), τp) is continuous. Let H ⊂ E be a bounded set. Then the ω∗ closure H is ω∗-compact, and then

~~τ ω∗ j(H) p = j(H ).~~

~~This combined with Theorem 4.11 yields~~

~~∗ ˆ τp ˆ ω d(j(H) ,C(BE )) = d(j(H ), C(BE )) = (4.23)~~

ˆ ω∗ sup d(j(z),C(BE )) = sup d(z,i(E)) = d(H , i(E)). (4.24) ω∗ ω∗ z∈H z∈H Therefore ∗ ˆ τp ˆ ω d(j(H) ,C(BE )) = d(H , i(E)). (4.25) For the next result, we refer to [7, Theorem 2.3].

Theorem 4.12 (a) For a bounded set H in a Banach space E, we have ck(H ) ≤ k(H) ≤ γ(H)≤ 2ck(H ) ≤ 2k(H ) ≤ 2ω(H), γ(H) = γ(conv(H )) and ω(H) = ω(conv(H )). 4.6 About compactness via distances to function spaces C(K) 133

∗ ∗∗ ω ∗∗ (b) For every x ∈ H , there exists a sequence (xn)n in H such that x − ∗∗ ∗∗ y ≤γ(H)for any cluster point y of (xn)n in E . Moreover, H is weakly relatively compact in E if and only if one (equivalently all) of the numbers ck(H ),k(H ), γ(H)or ω(H) equals zero.

Proof First observe that k(H) ≤ ε if Hε-interchanges limits with BE (see [7, Proposition 2.1(i)]). Hence k(H) ≤ γ(H). By Lemma 4.6, we deduce that γ(H)≤ 2ck(H ). The equality γ(H)= γ(conv(H )) follows from Theorem 4.8. The equality ω(H) = ω(conv(H )) is a consequence of the deﬁnition of ω(H) and the well-known fact stating that the closed convex hull of a weakly compact set in E is weakly compact; see [213, Theorem 9.8.5]. The last inequality is also easy. Indeed, take ε>0 and a weakly compact set Kε ⊂ E such that H ⊂ Kε + εBE. Then

~~ω∗ H ⊂ Kε + εBE ⊂ E + εBE .~~

Applying (4.20), we note k(H) ≤ ω(H). To prove the ﬁrst part of (b), it is enough ω∗ to use Theorem 4.10 (note that (H ,ω∗) can be looked at as a suitable subspace of B ([−M,M] E ,τp), where M is a bound of H ). Note also that ω(H) = 0 if and only if H is weakly relatively compact; see, for example, [123, Lemma 2, p. 227]. This completes the proof.

~~Finally, Theorem 4.12 ﬁxes the following well-known corollary.~~

~~Corollary 4.12 Let E be a normed space. Then (E, σ (E, E)) is angelic.~~

Proof We may assume that E is a Banach space. Let H be a weakly relatively countably compact set in E. Hence every sequence in H has a cluster point in σ(E,E), and this shows that ck(H ) = 0. By Theorem 4.12, H is weakly relatively compact. Fix x ∈ H , where the closure is taken in σ(E,E). Then γ(H)= 0 by Theorem 4.12, and we can apply (b) to get a sequence (xn)n in H such that

~~0 ≤y − x≤γ(H)= 0~~

~~for every σ(E,E )-cluster point y ∈ E of (xn)n. This implies xn → x in σ(E,E ) (since H is σ(E,E)-relatively compact and x is the unique σ(E,E)-cluster point of (xn)n).~~

~~We apply previous results to show Theorem 4.13 proved in [7, Theorem 3.5]. We need the following lemma; see [7].~~

Lemma 4.8 Let D ⊂ K be a dense subset of a compact space K, and let H ⊂ C(K) be a uniformly bounded set. If Hε-interchanges limits with D, then H 2ε- interchanges limits with K. 134 4 Web-Compact Spaces and Angelic Theorems

~~Proof Let δ>εbe arbitrary. We claim that, if f ∈ H (the closure in RK ), for each y ∈ K there exists a neighborhood V of y such that~~

~~sup |f(d)− f(y)|≤δ. d∈V ∩D~~

~~Indeed, assume that there exists y ∈ K such that~~

~~sup |f(d)− f(y)| >δ d∈U∩D for each neighborhood U of y. By a simple induction, we construct two sequences (gn)n in H and (dn)n in D such that~~

~~lim lim gm(dn) = lim f(dn), n m n~~

~~lim lim gm(dn) = lim gm(d0) = f(d0) = f(y). m n m Hence lim lim gm(dn) − lim lim gm(dn) = lim f(dn) − f(y) ≥ δ>ε. m n n m n~~

This provides a contradiction to the assumption that Hε-interchanges limits with D. The claim is proved. Next, choose arbitrary sequences (xn)n in K (with a cluster point x ∈ K) and (fm)m in H (with a cluster point f ∈ H ) such that the limits limn limm fm(xn) and limm limn fm(xn) exist. Then

~~lim lim fm(xn) = lim fm(x) = f(x), m n m~~

~~lim lim fm(xn) = lim f(xn). n m n Consequently, lim lim fm(xn) − lim lim fm(xn) = lim f(xn) − f(x) = A. m n n m n~~

We show that A ≤ 2δ. Since δ>ε, the proof will be finished. By the first part of the | − |≤ proof, we find a neighborhood U of x such that supd∈U∩D f(x) f(d) δ. Since 4.6 About compactness via distances to function spaces C(K) 135 for each n ∈ N there exists k>nsuch that xk ∈ U, the same argument is applied to get a neighborhood Vk ⊂ U of xk such that

~~sup |f(xk) − f(d)|≤δ. d∈Vk∩D~~

~~Take dk ∈ Vk ∩ D. Then~~

~~|f(xk) − f(x)|≤|f(xk) − f(dk)|+|f(dk) − f(x)|≤2δ.~~

~~Finally, we prove the following [7, Theorem 3.5].~~

~~Theorem 4.13 Let K be a compact space. Let H be a uniformly bounded set in C(K). Then γK (H ) ≤ γ(H)≤ 2γK (H ).~~

Proof Let M>0 be a uniform bound of H . For each x ∈ K,letδx : C(K) → R ∗ be the Dirac measure at x. Set D := {±δx : x ∈ K}. Note that conv(D) is ω -dense in BC(K) . If we show that HγK (H )-interchanges limits with conv(D), we can apply Lemma 4.8 to deduce that H 2γK (H )-interchanges limits with BC(K) , which H means γ(H)≤ 2γK (H ). To get this, note that D|H ⊂[−M,M] . Hence D|H γK (H )-interchanges limits with H . By Theorem 4.8, we deduce that conv(D)|H γK (H )-interchanges limits with H . Therefore HγK (H )-interchanges limits with conv(D).

~~Theorem 4.13 applies to extend Grothendieck’s characterization of weakly compact sets in Banach spaces; see [7, Corollary 3.6].~~

~~Corollary 4.13 Let K be a compact space. Then a uniformly bounded set H ⊂ C(K) is τp-relatively compact if and only if H is weakly relatively compact.~~

~~Proof By Corollary 4.8,thesetH is τp-relatively compact if and only if γK (H ) = 0. We apply Theorem 4.13. Chapter 5 Strongly Web-Compact Spaces and a Closed Graph Theorem~~

Abstract In this chapter, we continue the study of web-compact spaces. A subclass of web-compact spaces called strongly web-compact, is introduced, and a closed graph theorem for such spaces is provided. We prove that an own product of a strongly web-compact space need not be web-compact.

~~5.1 Strongly web-compact spaces~~

In this section, we introduce the class of strongly web-compact spaces [161]. A space X will be called strongly web-compact if X admits a family {Aα : α ∈ NN} of subsets of X (called a representation of X) covering X and such that, for = ∈ NN ∈ ∈ N every α (nk) ,ifxk Cn1,n2,...,nk for all k , the sequence (xk)k has a cluster point in X. Clearly, every strongly web-compact space is web-compact. We start with the following simple characterization.

~~Proposition 5.1 A space X is strongly web-compact if and only if X admits a N resolution {Aα : α ∈ N } of relatively countably compact sets.~~

N Proof If X is strongly web-compact, there exists a representation {Bα : α ∈ N } of X that is a resolution of relatively countably compact sets. Conversely, if X admits N a resolution {Aα : α ∈ N } of relatively countably compact sets covering X, then X N is strongly web-compact. Indeed, let α = (nk) ∈ N , and assume that ∈ := { : | = ∈ NN} xk Cn1,n2,...,nk Aα α k (n1,n2,...,nk), α

∈ N = k ∈ NN ∈ = k for all k . Then there exists βk (mn)n such that xk Aβk , nj mj ,for j = 1, 2,...,k.Let = { k : ∈ N} an max mn k ∈ N = ≥ ∈ N ⊂ ∈ for n and γ (an). Since γ βk for every k , Aβk Aγ . Hence xk Aγ for all k ∈ N. By the assumption, the sequence (xk)k has a cluster point in X.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 137 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_5, © Springer Science+Business Media, LLC 2011 138 5 Strongly Web-Compact Spaces and a Closed Graph Theorem

~~5.2 Products of strongly web-compact spaces~~

We provide an example from [161] showing that the square of a strongly web- compact space need not be strongly web-compact. Our approach uses some argu- mentsof[181, 9.15 Example] and some ideas presented in Novák’s example [314, Theorem 4].

~~Example 5.1 There exists a countably compact topological space G such that the N product G×G cannot be covered by a resolution {Aα : α ∈ N } of relatively countably compact sets.~~

Proof Let X be a discrete space of the cardinality c, and let X1 and X2 be two subspaces of X such that (i) X1 ∩ X2 =∅, (ii) X1 ∪ X2 = X, (iii) |X1| = |X2| = c. By (iii) there exists a bijection σ from X1 onto X2. Then its StoneÐCechˇ exten- β sion σ is a homeomorphism from βX1 onto βX2. Since X is a discrete space, we have βX βX X1 ∩ X2 =∅ and βX βX βX X1 ∪ X2 = X ; βX this follows from [146, 3.6.2]. If Y is a subspace of X, we can identify βY with Y ; see again [146, 3.6.8]. So, it follows that βX1 ∩ βX2 =∅and βX1 ∪ βX2 = βX. Moreover, if N is a countable inﬁnite subspace of X, βX N = |βN| = |βN| = 2c.

β Now define a homeomorphism ϕ : βX → βX by ϕ (x) = σ (x) if x ∈ βX1 and β −1 ϕ(x) = (σ ) (x) if x ∈ βX2. Clearly, ϕ (ϕ (p)) = p for every p ∈ βX, and p ∈ X if and only if ϕ (p) ∈ X. Since ϕ(βX1) = βX2 and ϕ(βX2) = βX1,themapϕ does not have fixed points. Set βX Z := N : N ∈ N , where N denotes the family of all countable infinite subsets of X.ByM we denote the family of all countable infinite subsets of Z. Since ℵ |N | = c 0 , we have ℵ |Z| = c 0 × 2c = 2c, 5.2 Products of strongly web-compact spaces 139 and hence |M | = 2c. So, if m is the first ordinal of cardinality 2c, we have that c M = {Mα : 0 ≤ α

{Ni : i ∈ N} ⊆ N with ∞ βX βX M ⊆ Ni . i=1 ∞ ∈ N N c Since i=1 Ni and every inﬁnite closed subset of β has cardinality 2 (see βX [146, 3.6.14]), |M |=2c. Let βX y0 ∈ M0 \ M0. βX Let 1 ≤ α

~~Recall that countable products of K-analytic spaces are also K-analytic. Exam- ple 5.1 yields the following corollary.~~

~~Corollary 5.1 There exists a quasi-Suslin space X such that X × X is not quasi- Suslin.~~

~~5.3 A closed graph theorem for strongly web-compact spaces~~

It turns out that the concept of strongly web-compact spaces can be used to extend some classical closed graph theorems; see [156]. Valdivia [421, I.4.2 (11)] proved that a linear map with a closed graph from a metrizable Baire lcs E into a quasi-Suslin lcs F is continuous. Drewnowski [131, Corollary 4.10] proved that every continuous linear map from a tvs having a compact resolution onto an F-space (i.e., a metrizable and complete tvs) is open. In this section, we use some techniques of [421] to get a closed graph theorem that extends Valdivia’s [421, I.4.2 (11)] and Drewnowski’s [131, Corollary 4.10, Corollary 4.11]. In Theorem 5.1, we need only to assume that F is a tvs with a relatively countably compact resolution and E is a Baire tvs. We start with some additional facts that will be used in this section. Fact 1. Let B be a subset of a topological space E. Assume B is covered by a { : ∈ N} web Bn1,n2,...,np p,n1,n2,...,np and that

⊂ O(Bn1,n2,...,np ) B p N for each sequence (np)p in . By Proposition 2.4,thesetB has the Baire property. Indeed, H := O(B)\ O(B ) and n1∈N n1 := \ Hn1,n2,...,np O(Bn1,n2,...,np ) O(Bn1,n2,...,np,m), m∈N where p,n1,n2,...,np ∈ N, are nowhere dense sets. By the hypothesis, ∪ : ∈ N H Hn1,n2,...,np p,n1,n2,...,np contains O(B)\B. Therefore the set B has the Baire property. Fact 2. Let E be a topological space admitting a weaker ﬁrst-countable topology N τ .IfB is a subset of E having a τ -closed resolution {Aα : α ∈ N },thesetB N ∈ ∈ N has the Baire property. Indeed, choose (np)p in and x O(Cn1,n2,...,np ), p , 5.3 A closed graph theorem for strongly web-compact spaces 141 := { : | = } where, as usual, Cn1,n2,...,np Aα α p (n1,n2,...,np) .Let(Up)p be a τ - ∈ ∩ ∈ NN neighborhood basis of x. Select xp Up Cn1,n2,...,np . There exists α such that xp ∈ Aα for p ∈ N. By the τ -closedness condition, we get that x ∈ Aα ⊂ B and Fact 1 applies. By F(E) we denote a basis of balanced neighbourhoods of zero in a tvs E.Now we prove the following closed graph theorem from [156].

Theorem 5.1 Let E and F be tvs such that E is Baire and F admits a relatively N countably compact resolution {Aα : α ∈ N }. If f : E → F is a linear map with a closed graph, there is a sequence (Un)n in F(E) such that for every V ∈ F(F ) there −1 −1 exists m ∈ N with m Um ⊂ f (V ). Hence f is continuous. If E = F , then E is a separable F-space.

~~Proof Since E is a Baire space, there exists a sequence (rp)p in N such that −1 −1 f (Hp) − f (Hp) is a neighborhood of zero for each p ∈ N, where Hp :=~~

~~Cr1,r2,...,rp .Let(Up)p be a sequence of balanced neighborhoods of zero in E such that~~

Up+1 + Up+1 ⊂ Up and −1 −1 Up ⊆ f Hp − f Hp for each p ∈ N. Let τ be the semimetrizable translation-invariant vector topology on −1 E deﬁned by the basis (p Up)p of neighborhoods of zero. Since the graph of f is closed, there is a coarser linear topology ρ on F such that the map f : E → (F,ρ) is continuous. We claim that f : (E,τ) → (F,ρ) is continuous. Indeed, if V is a closed neighborhood of zero in (F,ρ), there exists q ∈ N such that −1 q (Hq − Hq ) ⊂ V. As f −1(V ) is closed in E,

−1 −1 −1 −1 −1 q Uq ⊂ q (f (Hq ) − f (Hq )) ⊂ f (V ). This proves the claim. If W ∈ F(F ) is closed and balanced, and −1 Bα := f (Aα ∩ W) , N then (using the fact that {Aα : α ∈ N } is relatively countably compact) we deduce τ −1 τ N −1 that Bα ⊂ f (W). Thus {Bα : α ∈ N } is a τ -closed resolution for f (W), and f −1 (W) has the Baire property by Fact 2. Since f −1 (W) is of second category, Proposition 2.7 ensures that f −1 (W) − f −1 (W) is a neighborhood of zero in E. Hence f : E → F is continuous. If V ∈ F(F ) is closed, there is m ∈ N such that −1 −1 m Hm − m Hm ⊂ V, 142 5 Strongly Web-Compact Spaces and a Closed Graph Theorem

−1 −1 so m Um ⊂ f (V ).IfE = F , then E is a metrizable tvs having a compact resolution, and Corollary 6.2 (below) is applied to deduce that (E, τ) is analytic and hence separable. Since every analytic Baire tvs is metrizable and complete (see The- orem 7.2 below), the proof is completed.

Theorem 5.1 fails for topological groups in general. If a compact group of Ulam measurable cardinality is either Abelian or connected, then it admits a strictly ﬁner countably compact group topology [101].

Corollary 5.2 Every linear map f from a Baire tvs E into a tvs F whose graph G admits a relatively countably compact resolution is continuous. Hence, every linear map from a metrizable and complete tvs into a separable metrizable tvs whose graph admits a complete resolution is continuous.

Proof The projection P(x,f(x)) = x of G onto E is continuous, so P −1 is continuous by Theorem 5.1. Note that f = Q ◦ P −1, where Q : G → F is the projection. To complete the proof, we show that f is continuous on every closed separable vec- N tor subspace E0 of E.Let{Aα : α ∈ N } be a complete resolution in G.Thesets Aα ∩ (E0 × F)form a complete resolution on G ∩ (E0 × F). By Corollary 6.2 (below), every metrizable and separable tvs having a complete resolution is analytic.

~~Corollary 5.3 is a special case of Theorem 7.4 below.~~

Corollary 5.3 Let f be a linear functional on a metrizable and complete tvs E. The following conditions are equivalent: N (i) E admits a resolution {Aα : α ∈ N } such that f is continuous on each Aα. (ii) f is continuous on E. (iii) The kernel N := {x ∈ E : f(x)= 0} has a complete resolution.

Proof (i) ⇒ (ii): Observe that f is continuous on the closure on each Aα. Indeed (we follow the argument from [131, Proposition 4.1]), ﬁx Aα and x ∈ Aα. There N exists β ∈ N such that Aα ∪{x}⊂Aβ . Since, by the assumption, f |Aβ is continuous at x, f |(Aα ∪{x}) is also continuous at x. This means that the limit of f |Aα at point x exists and equals f(x). Hence f |Aα is continuous. We may assume that each set Aα is closed. Also, we may assume that E is separable. As every metrizable and complete separable tvs admits a compact resolution, ﬁx a compact N resolution {Kα : α ∈ N } on E. Then Dα := Aα ∩ Kα composes a compact resolution in E, and f is continuous on each Dα. Assume that f is discontinuous, and N let H be its (dense) kernel. Clearly, {H ∩ Dα : α ∈ N } is a compact resolution on H .Also,H admits a strictly weaker metrizable and complete vector topology. In- deed, if D is an algebraic complement to H in E, the restriction of the quotient map q|H : H → E/D generates such a topology on H . Theorem 5.1 is applied to reach a contradiction. A similar argument as in (i) ⇒ (ii) applies to get (iii) ⇒ (i). Chapter 6 Weakly Analytic Spaces

Abstract This chapter studies analytic spaces. We show that a regular space X is analytic if and only if it has a compact resolution and admits a weaker metric topology. This fact, essentially due to Talagrand, extends Choquet’s theorem (every metric K-analytic space is analytic). Several applications will be provided. We show Christensen’s theorem stating that a separable metric topological space X is a Polish space if and only if it admits a compact resolution swallowing compact sets. We also study the following general problem: When can analyticity or K-analyticity of the weak topology σ(E,E) of a dual pair (E, E) be lifted to stronger topologies on E compatible with the dual pair? We prove that, if X is an uncountable analytic space, the Mackey duals Lμ(X) of Cp(X) is weakly analytic and not analytic. The density condition, due to Heinrich, motivates us to study the analyticity of the Mackey and strong duals of (LF )-spaces. We study trans-separable spaces and show that a tvs with a resolution of precompact sets is trans-separable. This is applied to prove that precompact sets are metrizable in any uniform space whose uniformity admits a U -basis.

~~6.1 A few facts about analytic spaces~~

In this section, we collect some general facts about analytic tvs’s. Recall again that a topological space E is analytic if E is a continuous image of a Polish space (or, equivalently, of the space NN).

Proposition 6.1 (i) Every open (closed) subspace of an analytic space E is analytic. (ii) Any countable product n En of analytic spaces is an analytic space. (iii) Every countable union (intersection) of analytic subspaces of a topological space is an analytic space. (iv) If E is an analytic space and F is its closed subspace, then the quotient E/F is an analytic space.

~~Proof (i) Let U be an open (closed) subspace of E, and let T : P → E be a continuous map from a Polish P space onto E. Then T −1(U) is a Polish subspace of E~~

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 143 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_6, © Springer Science+Business Media, LLC 2011 144 6 Weakly Analytic Spaces by Proposition 2.8.ThemapT |T −1(U) : T −1(U) → U is a continuous surjection from a Polish space T −1(U). Hence U is analytic. ∈ N : → (ii) For each n ,letPnbe a Polish space, and let Tn Pn En be a continuous = : → map onto En. Clearly, P n Pn is a Polish space and the mapT P n En de- := ﬁned by T ((xn)) (Tn(xn)) is a continuous surjection onto n En. Hence n En is analytic. The remaining claims are left to the reader.

~~Proposition 6.2 Every Borel subset of an analytic space E is analytic.~~

Proof Set U := {A ⊂ E : A, E \ A are analytic}. Note that U contains all open subsets of E. It is enough to prove that U is an σ -algebra in E. The proof will U be completed if we realize that if (An)n is a sequence of subsets of , then ∈ U n An .

~~To prove Proposition 6.3, we need the following simple lemma; see [146].~~

~~Lemma 6.1 A Lindelöf regular topological space X is normal.~~

~~Proof Let A and B be two disjoint closed subsets of X. For points a ∈ A and b ∈ B, let Ua and Vb be open neighborhoods of a and b, respectively, such that Ua ∩ B =∅ and Vb ∩ A =∅. Clearly,~~

{Ua : a ∈ A} ∪ {Vb : b ∈ B} ∪ {X\(A ∪ B)} is an open cover of X. By the assumption on X, there exist a sequence (Un)n in { : ∈ } { : ∈ } ⊂ ⊂ Ua a A and a sequence (Vn)n in Vb b B such that A n Un and B n Vn. Set ∗ = \ : ≤ Un Un Vm m n and ∗ = \ : ≤ Vn Vn Um m n . = ∗ = ∗ ⊂ ⊂ Then open sets U n Un and V n Vn are disjoint, A U and B V . Hence X is normal.

The following applicable result was obtained by Talagrand [392]. Proposition 6.3 extends Choquet’s theorem from [97] (every metric K-analytic space is analytic). The proof presented below is a modiﬁcation of the proof due to Cascales and Oncina; see [85, Corollary 4.3] and also [346, Theorem 5.5.1], [371, Corollary 1, p. 105].

Proposition 6.3 (Talagrand) Let (X, τ) be a K-analytic space. Let d be a metric on X whose topology is coarser than τ . Then (X, τ) is analytic. Every regular analytic space X admits a weaker metric topology. 6.1 A few facts about analytic spaces 145

N Proof Let {Kα : α ∈ N } be a compact resolution on (X, τ), and let {zn : n ∈ N} be a dense subset of (X, d).ByBd (z, r) denote the d-closed ball in (X, d) of center z N and radius r>0. For β = (bn) ∈ N ,let := −1 Dβ Bd (zbn ,n ). n∈N

N N Each set Dβ is unitary or void. For y ∈ X, there exists (α, β) ∈ N × N such that Kα ∩ Dβ = {y}. For Kα ∩Dβ = ∅, we denote by yαβ the element of X such that Kα ∩Dβ ={yαβ }. If N N T := (α, β) ∈ N × N :∅ =Kα ∩ Dβ = yαβ , the map f : T → X deﬁned by f ((α, β)) = yαβ is surjective. NN × NN Let (α(p),β(p))p be a sequence in T that converges to (α, β) in , and let (α(p),β(p))p(m) be a subsequence. By the K-analyticity, (yα(p),β(p))p(m) has an adherent point y ∈ Kα. Since β(p) N converges to β = (bn)n ∈ N , the sequence (yα(p),β(p))p(m) is eventually in each −1 −1 Bd (zbn ,n ), and hence its adherent point y belongs to Bd (zbn ,n ). This shows that y ∈ Kα ∩ Dβ = yαβ . We proved that (α, β) ∈ T (i.e., T is a closed subset of NN ×NN and therefore a Pol- ish space). Moreover, we proved that yαβ is an adherent point of each subsequence of (yα(p),β(p))p. This implies that yα(p),β(p) converges to yαβ (i.e., f (α(p),β(p)) converges to f (α, β)). Hence f is a continuous mapping from the Polish space T onto (Y, τ), and this proves that (Y, τ) is analytic. In order to prove the second part of the proposition, let Δ = {(x, x) : x ∈ X} be the diagonal of the analytic space X × X. Clearly, Δ and (X × X)\Δ are analytic, and therefore they are Lindelöf. If x = y, there exist two closed neighborhoods Fx and Fy of x and y, respectively, such that

~~Fx × Fy ⊂ (X × X)\Δ.~~

~~The Lindelöf property enables us to determine a sequence (xn,yn)n such that × \ = × X X Δ Fxn Fyn . n~~

~~× = Therefore Δ is a Gδ-subset of X X since Δ n Gn, where = × \ × Gn (X X) (Fxn Fyn ).~~

~~For each (x, x) ∈ Δ and n ∈ N, there exists an open set Ux,n in X such that~~

~~(x, x) ∈ Ux,n × Ux,n ⊂ Gn. 146 6 Weakly Analytic Spaces~~

~~As the space X is completely regular by Lemma 6.1, we may assume that there exists a continuous function fx,n : X → [0, 1] such that~~

~~1 1 fx,n(Ux,n) ⊂ , 1 ,fx,n(X\Ux,n) ⊂ 0, . 2 2~~

~~By the Lindelöf property of Δ, the family {Ux,n : x ∈ X} contains a sequence~~

~~(Ux(i,n),n)i such that ⊂ × := ∗ Δ Ux(i,n),n Ux(i,n),n Gn. i~~

= ∗ = Since Δ n Gn, we deduce that, if x y are two different points of X, there ∈ N ∈ ∗ ∈ ∗ ∈ N exists n such that (x, y) / Gn. Then, since (x, x) Gn, there exists j such ∈ ∈ ∈ ∗ that x Ux(j,n),n. This implies that y/Ux(j,n),n since (x, y) / Gn. By the construction, = fx(j,n),n(x) fx(j,n),n(y). Then, X endowed with the topology that makes the countable family of functions { : ∈ N2} fx(i,n),n (i, n) continuous is metrizable with the metric deﬁned by the formula = −i−n − : ∈ N2 d(x,y) 2 fx(i,n),n(x) fx(i,n),n(y) (i, n) . Clearly, d(x,y) deﬁnes a metric topology weaker then τ .

~~Corollary 6.1 A compact space is analytic if and only if it is metrizable.~~

Note that every separable and complete metric space is a continuous and open image of NN;see[346, Part 3, Theorem 1.2.14] and Theorem 3.9. In order to ﬁx another sufﬁcient and necessary condition for a K-analytic space to be analytic (see [346, Theorem 5.5.1]), we need the following simple lemma.

Lemma 6.2 Suppose that U is an open subset of a topological space X. If there ≤ ≤ n ⊂ exist compact subsets Ki ,1 i n, such that i=1 Ki U, then there exist neigh- ≤ ≤ n ⊂ borhoods Ui of Ki ,1 i n, such that i=1 Ui U.

Proof For n = 2, the lemma follows from the well-known fact stating that two disjoint compact subsets have disjoint open neighborhoods. Therefore, there exist open \ = ∩ =∅ := neighborhoods Vi of Ki U, i 1, 2, such that V1 V2 . Then the sets Ui ∪ = n ⊂ U Vi , i 1, 2, are as desired. Therefore, if i=1 Ki U, then there exist two n−1 ⊂ ⊂ ∩ ⊂ open subsets Vn−1 and Un such that i=1 Ki Vn−1,Kn Un,Vn−1 Un U. Then, if the claim is true for n − 1, it is also true for n. Now the lemma follows by a simple induction.

~~We prove a part of [346, Theorem 5.5.1]. 6.1 A few facts about analytic spaces 147~~

~~Proposition 6.4 Let X be a K-analytic space. The following statements are equivalent: (a) X is analytic. (b) X is a continuous image of a separable metric space.~~

Proof Clearly (a) implies (b). Therefore, we only need to prove that (b) implies (a). If X is a continuous image of a separable metric space, X × X is also a continuous image of a separable metric space. The space X × X is hereditarily Lindelöf. There- fore, if Δ := {(x, x) : x ∈ X}, then (X × X)\Δ is an open Lindelöf subset of X × X, 1 2 and then there exists a sequence (Gn,Gn)n of pairs of open sets such that × \ = 1 × 2 (X X) Δ Gn Gn. (6.1) n

For each n ∈ N,let i := \ i = An X Gn,i 1, 2. (6.2) i By the K-analyticity of X, we deduce that each closed set An is K-analytic, and then N i i : N → An there exists a compact-valued usco map Kn 2 yielding the K-analyticity i ∈ N of An. Clearly, for each n , the compact-valued map

N X Ln :{1, 2}×N → 2 deﬁned by := i Ln((i, ω)) Kn(ω) is also usco. Let L be the map from ({1, 2}×NN)N into 2X deﬁned by L((i(n), ω(n))n) := Ln((i(n), ω(n))), n where each i(n) ∈{1, 2}, ω(n) ∈ NN. Then = i(n) L((i(n), ω(n))n) Kn (ω(n)). (6.3) n

~~i(n) ⊂ Let U be an open subset of X such that n Kn (ω(n)) U. By the compactness, there exists n0 such that~~

n0 i(m) ⊂ Km (ω(m)) U. m=1 ≤ ≤ Lemma 6.2 implies that for each 1 m n0 there exists a neighborhood Um of i(m) n0 ⊂ Km (ω(m)) such that m=1 Um U. Then, by the upper semicontinuity of each 148 6 Weakly Analytic Spaces

~~i(m) i(m) Km , there exists a neighborhood Vm of ω(m) such that Km (Vm) ⊂ Um for each 1 ≤ m ≤ n0. Therefore n0 i(m) ⊂ Km (Vm) U. m=1 Hence, for the open set~~

~~N N V := {(i(n), ω(n))n ∈ ({1, 2}×N ) : i(m) = im,ω(m)∈ Vm : 1 ≤ m ≤ n0}, we note that L(V ) ⊂ U. For U =∅,theset~~

~~N N {(i(n), ω(n))n ∈ ({1, 2}×N ) : L((i(n), ω(n))n) =∅} is open, and then~~

N N F := {(i(n), ω(n))n ∈ ({1, 2}×N ) : L((i(n), ω(n))n) = ∅} is a closed subset of the separable complete and metrizable space ({1, 2}×NN)N. The restriction of L to the Polish space F is a compact-valued usco map. In order to complete the proof, we only need to show that L(F ) = X and the compact values of L|F are unitary sets. ∈ N 1 ∪ 2 = From (6.1) and (6.2), it follows that for each n we have An An X, and this clearly implies that L(F ) = X. Finally, we prove that if (i(n), ω(n))n ∈ F , then L((i(n), ω(n))n) is a unitary set. Indeed, from (6.1) we deduce that for x = y there ∈ 1 × 2 ∈ 1 ∈ 2 exists q such that (x, y) Gq Gq . This and (6.2) yield x/Aq and y/Aq . This { } i = i ⊂ i implies that x,y Aq for i 1, 2. From Kq (ω) Aq , it follows that

~~{ } i x,y Kq (ω) and, in particular, i(q) {x,y} Kq (ω(q)). (6.4) Then, from (6.3) and (6.4), we deduce~~

~~{x,y} L((i(n), ω(n))n).~~

This proves that for (i(n), ω(n))n ⊂ F and different points x and y in X the set {x,y} is not contained in L((i(n), ω(n))n). We proved that the values of L|F are compact unitary sets and the map f : F → X deﬁned by

~~{f ((i(n), ω(n))n)}=L((i(n), ω(n))n) is a continuous surjection. We showed that (b) ⇒ (a) since F is a Polish space.~~

~~Proposition 6.5 Let E be a separable lcs. Then (E,σ(E,E)) is strongly web- compact if and only if (E,σ(E,E))is analytic. 6.2 Christensen’s theorem 149~~

Proof Clearly, every analytic space is strongly web-compact. For the converse, assume (E,σ(E,E)) is a strongly web-compact space. Since E is separable, there exists on E a metric topology ξ weaker than the topology σ(E,E). Hence (E,σ(E,E)) is angelic. Consequently, every relatively countably compact set in σ(E,E) is relatively compact. Therefore (E,σ(E,E)) admits a compact resolution. Then, by Corollary 3.6, the space (E,σ(E,E)) is K-analytic. Now Proposi- tion 6.3 applies.

~~6.2 Christensen’s theorem~~

We know already that every Polish space E admits a compact resolution {Kα : α ∈ NN} swallowing compact sets (i.e., every compact set in E is contained in some Kα). Clearly, every hemicompact space E with a fundamental (increasing) sequence (Kn)n of compact sets generates a compact resolution swallowing compact sets. := = ∈ NN Q Indeed, it is enough to set Kα Kn1 for any α (nk) . The space of the rational numbers is analytic and does not admit a compact resolution swallowing compact sets; this follows from Theorem 6.1 below since Q is not complete. Clearly, every σ -compact space is Lindelöf. Nevertheless, there exist locally compact spaces having a compact resolution swallowing compact sets that are not Lindelöf. The following interesting example is due to Tkachuk [399].

~~Example 6.1 (Tkachuk) There exists a locally compact space E that is a countably compact and noncompact space. Moreover, E has a compact resolution swallowing compact sets.~~

~~Proof Let NN be endowed with the discrete topology. For any α ∈ NN,set~~

N ∗ Aα := {β ∈ N : β ≤ α}, ≤∗ = ∈ N where the relation β = (bn) α (an) means that there exists m such that ≤ ≥ ⊂ ≤ := bn an for all n m. Clearly, Aα Aβ for α β. Set E α Kα, where N N N Kα := Aα and closure is taken in βN . Hence N ⊂ E ⊂ βN , and E is locally compact, having a compact resolution of open subsets of βNN. Note that E is countably compact. Indeed, choose an arbitrary countable set A ⊂ E. Then there exists a ⊂ NN ⊂ ∈ NN ≤∗ countable set D such that A α∈D Kα. Choose β such that α β for any α ∈ D. Then A ⊂ Kα ⊂ Kβ ⊂ E. α∈D Hence, the closure of every countable subset of E is compact and therefore E is N countably compact. Note also that the resolution {Kα : α ∈ N } swallows compact sets. Indeed, the sets Kα are also open sets, and if K ⊂ E is a compact set in E, 150 6 Weakly Analytic Spaces ={ }⊂NN ⊂ there exists a finite set C α1,α2,...,αk such that K α∈C Kα. Set := k j := j = bn j=1 an, where αj (an).Letβ (bn). Then K ⊂ Kα ⊂ Kβ . α∈C Next, observe that E is noncompact and hence, being countably compact, it cannot N N be Lindelöf. Indeed, set F ={Wα : α ∈ N }, where Wα := N \ Aα. The family F ={ } has the finite intersection property. To see this, for a finite set B α1,α2,...,αk NN := k j + := j ≤ ≤ := in ,set bn j=1 an 1, where αj (an)n for 1 j k.Ifβ (bn),we ∈ have β α∈B Wα. Consequently, U := Wα = ∅. α

~~N N N Since Kα ∩ Wα =∅for each α ∈ N ,wehaveU ⊂ βN \ E.AsE = βN ,we conclude that E is noncompact.~~

Theorem 6.1 below, due to Christensen [99, Theorem 3.3], also provides a good motivation to study spaces having a compact resolution swallowing compact sets. First we recall some typical notation. Following Suslin schemes notation (see Sec- tion 3.4), for N σ = (σ1,σ2,...,σn,...)∈ N 0 set σ |0 := ∅, and for n ∈ N set σ |n := (σ1,σ2,...,σn). Let N := ∅, and let N(N) := {Nn : n ∈ N}={σ |n : σ ∈ NN,n∈ N}.

~~By B(x,r) we denote (as usual) the open ball with the center x and radius r for a metrizable space X. We need two claims.~~

~~N Lemma 6.3 If a metrizable space X admits a compact resolution {Kσ : σ ∈ N } swallowing compact subsets of X, there exists an open covering~~

~~{A((σ1)), σ1 ∈ N} of X and a map q : N → N such that for each compact subset K of A((σ1)) there N exists α ∈ N such that α|1 = q((σ1)) and K ⊂ Kα.~~

Proof First we prove that for each x ∈ X there exists nx ∈ N such that for each −1 ∈ NN | = ⊂ compact subset K of B(x,nx ) there exists α such that α 1 (nx) and K Kα. Assume the contrary. Then there exists x ∈ X such that for each n ∈ N there −1 N exists a compact subset Kn of X such that Kn ⊂ B(x,n ) and for each α ∈ N with α|1 = (n)

~~Kn Kα. (6.5) 6.2 Christensen’s theorem 151 := { }∪ Clearly, K x n Kn is a compact subset of X. By the assumption, there exists α ∈ NN such that~~

~~K ⊂ Kα. (6.6)~~

If α|1 = (m), then (6.6) implies Km ⊂ Kα, which provides a contradiction with = { −1 : n m. This proves the claim. Since X is Lindelöf, the open covering B(x,nx ) x ∈ X} admits a countable subfamily

~~−1 {B(xσ ,n ) : σ1 ∈ N} 1 xσ1 covering X. The lemma is proved with −1 A((σ1)) = B xσ ,n 1 xσ1 and q : N → N, which is deﬁned by q((σ )) = (n ). 1 xσ1~~

Clearly, we may apply Lemma 6.3 to each of the sets A((σ1)). Then we can obtain a covering {A((σ1,σ2)), σ2 ∈ N} of each of the sets A((σ1)) constructed in Lemma 6.3. This process can be continued by induction. Next, Lemma 6.4 describes this construction.

Lemma 6.4 Let X be a metrizable topological space that admits a compact reso- N lution {Kσ : σ ∈ N } swallowing compact sets of X. Let A((σ1,σ2,..., σh)) be an h open subset of X, and let q((σ1,σ2,..., σh)) be a multiindex of N such that for N each compact subset K of A((σ1,σ2,..., σh)) there exists α ∈ N such that

~~α|h = q((σ1,σ2,..., σh)) and K ⊂ Kα. Then there exists an open covering~~

~~{A((σ1,σ2,..., σh,σh+1)) : σh+1 ∈ N}~~

1 1 of A(σ1,σ2,..., σh), and a map q :{(σ1,σ2,..., σh)}×N → N such that for N each compact subset K of A((σ1,σ2,..., σh,σh+1)) there exists α ∈ N such that α|(h + 1) = (q((σ1,σ2,..., σh)), q ((σ1,σ2,..., σh,σh+1))) and K ⊂ Kα.

~~Proof As X is a Lindelöf space, we need only to prove that for each~~

~~x ∈ A(σ1,σ2,..., σh)~~

−1 there exists n ∈ N such that B(x,n ) ⊂ A(σ1,σ2,..., σh) and for each compact N set K ⊂ B(x,n) there exists α ∈ N such that α|h + 1 = (q((σ1,σ2,..., σh)), n) and K ⊂ Kα. Assume the contrary. Then there exists x ∈ A(σ1,σ2,..., σh), and let −1 ⊂ n0 be the minimum natural number such that B(x,n0 ) A(σ1,σ2,..., σh). Then, 152 6 Weakly Analytic Spaces

~~−1 for each n = n0,n0 + 1,..., there exists a compact subset Kn of B(x,n ) such that for each α ∈ NN with~~

~~α|h + 1 = (q((σ1,σ2,..., σh)), n) we have~~

~~Kn Kα. (6.7) := { }∪[ ] Clearly, K x n Kn is a compact subset of A(σ1,σ2,..., σh). By the as- N sumption, there exists α ∈ N ,α|h = q((σ1,σ2,..., σh)), such that~~

~~K ⊂ Kα. (6.8)~~

~~Then, Kp ⊂ Kα for α|h + 1 = (q((σ1,σ2,..., σh)), p). This contradicts (6.7)for n = p.Themapq is deﬁned similarly to Lemma 6.3.~~

~~Now we prove the following deep result due to Christensen [99].~~

~~Theorem 6.1 (Christensen) If a metrizable topological space X admits a compact N resolution {Kσ : σ ∈ N } swallowing compact subsets, then X is a Polish space.~~

~~Proof Using Lemmas 6.3 and 6.4, we construct a Suslin scheme A(·) of nonempty open subsets of X and a map q : N(N) → N(N) such that A(∅) = X = {A(σ1) : σ1 ∈ N}. (6.9)~~

~~k For each k ∈ N and each (σ1,σ2,...,σk) ∈ N ,wehave A((σ1,σ2,...,σk)) = {A((σ1,σ2,...,σk,σk+1)) : σk+1 ∈ N}, (6.10)~~

~~k q(∅) =∅, and for each k ∈ N and each (σ1,σ2,...,σk) ∈ N~~

~~q((σ1,σ2,...,σk))|i = q((σ1,σ2,...,σi)) (6.11)~~

~~k for 1 ≤ i ≤ k − 1. For each (σ1,σ2,...,σk) ∈ N and any compact set~~

~~K ⊂ A((σ1,σ2,...,σk)), there exists α ∈ NN such that~~

~~α|k = q((σ1,σ2,...,σk)) (6.12) and~~

~~K ⊂ Kα. (6.13) 6.2 Christensen’s theorem 153~~

~~Let d be a metric on X, and let X be the d-completion of X. For each (σ1,...,σk),let~~

~~ X B((σ1,σ2,...,σk)) := X\{X\A((σ1,σ2,...,σk)) }.~~

~~Since A((σ1,σ2,...,σk)) is an open subset of X,~~

~~X {X\A((σ1,σ2,...,σk)) }∩X = X\A((σ1,σ2,...,σk)).~~

~~Hence~~

~~B((σ1,σ2,...,σk)) ∩ X = A((σ1,σ2,...,σk)). Let N M := {B((σ1,σ2,...,σk)), k ∈ N},σ = (σ1,...)∈ N . (6.14)~~

~~Note that X ⊂ M by (6.12) and (6.10). For each y ∈ M, we may choose σ = N (σ1,σ2,...)∈ N such that y ∈ B((σ1,σ2,...,σk)) for each k ∈ N. Then, for each k ∈ N, there exists~~

~~xk ∈ A((σ1,σ2,...,σk)) (6.15) such that −1 d(y,xk)~~

~~α(k)|k = q((σ1,σ2,...,σk))~~

N and {xk}⊂Kα(k). Then β := sup{α(k) : k ∈ N}∈N by (6.11). Since {xk : k ∈ N}⊂ Kβ , the sequence (xk)k has an adherent point in X. Then (6.16) implies that y ∈ X. From (6.14), it follows that X\M is the union of the sets X [B((σ1)) : σ1 ∈ N] , (6.17)

~~ B((σ1,σ2,...,σk)) [B((σ1,σ2,...,σk,σk+1)) : σk+1 ∈ N] (6.18)~~

k for each k ∈ N, (σ1,σ2,...,σk) ∈ N . By the separability of X, we note that each set in (6.18) is a countable union of closed sets. Since the set in (6.17)isclosed,X\M is an Fσ -subset of X, and then M = X is a Gδ-subset of X. Hence X is Polish as a Gδ-subset of the Polish space X.

~~Proposition 6.3 is applied to provide the following characterization of the analyticity for metric spaces. 154 6 Weakly Analytic Spaces~~

~~Corollary 6.2 For a metric space X, the following assertions are equivalent: (i) X is separable and admits a complete resolution. (ii) X admits a compact resolution. (iii) X is analytic.~~

N N Proof (i) ⇒ (ii): Let {Bα : α ∈ N } be a complete resolution, and let {Tα : α ∈ N } be a compact resolution on the completion Y of X. Then the sets Kα := Tα ∩ Bα form a compact resolution on X for α ∈ NN. (ii) ⇒ (iii) follows from Proposition 6.3. (iii) ⇒ (i) since every analytic space is separable and admits a compact resolution.

Corollary 6.2 may suggest the following problem. Let X be a Banach space that is weakly K-analytic. Assume that every weakly compact set is separable. Is E separable? The answer is positive under Martin’s axiom plus the negation of the CH (see [171]). It fails under the CH: see, for example, [349]. Mercourakis and Stamati [295] introduced a class of Banach spaces E (under the name strongly weakly K-analytic (SWKA) whose weak topology σ(E,E) admits an usco map T : NN → K (E), where K (E) is the family of all σ(E,E)-compact sets in E and such that for every compact set K in (E, σ (E, E )) there exists Tα := T(α)such that K ⊂ Tα. On the other hand, if E is a metrizable lcs whose weak topology σ(E,E) ad- { : ∈ NN} mits a compact resolution Kα α swallowing compact sets, then a map : : NN → K := T T (E) deﬁned by T(α) k Cn1,n2,...,nk , is usco for the weak topology σ(E,E ). Indeed, since (E, σ (E, E )) is angelic from Corollary 4.3, countable compact sets k Cn1,n2,...,nk are compact; see Corollary 3.6. Clearly, N Kα ⊂ Cn ,n ,...,n for each k ∈ N. We provided an usco map T : N → K (E) for 1 2 k the weak topology σ(E,E ) such that for every K in (E, σ (E, E )) there exists Tα such that K ⊂ Tα. For a Banach space E,letB(E) be the unit closed ball in E endowed with the weak topology σ(E,E).IfB(E) is a Polish space, then it is metrizable, and it is well known that the weak∗ dual E of E is separable. It is also a classical fact (see [183]) that, for a Banach space E such that E is separable, the ball B(E) with the weak topology of E is a Polish space. An application of Theorem 6.1 shows the following; see [295].

~~Proposition 6.6 Let E be a Banach space whose dual E is separable. Then E is SWKA if and only if the closed unit ball B(E) is a Polish space in the weak topology of E.~~

Proof Since E is separable, B(E) is σ(E,E)-metrizable. Assume E is SWKA. Since B(E) admits a compact resolution swallowing compact sets in the topology σ(E,E)|B(E), we apply Theorem 6.1 to complete the proof for this case. Con- versely, if B(E) is a Polish space, it is K-analytic and admits a compact resolution swallowing compact sets. 6.3 Subspaces of analytic spaces 155

Since for the Banach space c0 the closed unit ball of c0 is not a Polish space in the weak topology, Proposition 6.6 shows that c0 is not SWKA. Since every closed subspace of an SWKA Banach space is SWKA, and for an inﬁnite metrizable compact space K the Banach space Cc(K) contains an isomorphic copy of c0,we note the following corollary.

~~Corollary 6.3 If K is an inﬁnite metrizable compact space, Cc(K) does not admit a compact resolution in the weak topology that swallows weakly compact sets.~~

~~On the other hand, Edgar and Wheeler proved [145, Theorems A, B] the following.~~

Theorem 6.2 Let E be a separable Banach space. The following assertions are equivalent: (i) B(E) endowed with the topology σ(E,E) is completely metrizable. (ii) B(E) endowed with the topology σ(E,E) is a Polish space. (iii) For every σ(E,E)-closed bounded set B ⊂ E, the identity map (B, σ (E, E)) → E has at least one point of continuity. (iv) B(E) endowed with the topology σ(E,E) is metrizable, and every closed subset of B(E) is a Baire space.

It is well known that every Cech-completeˇ space X (i.e., X is a Gδ-set in some (every) compactiﬁcation of X) is a Baire space; see [146]. If E is an inﬁnite- dimensional Banach space, then (E, σ (E, E)) is not a Baire space, although B(E) is Cech-completeˇ in the weak topology [145].

~~6.3 Subspaces of analytic spaces~~

In this section, we provide some applications of Theorem 5.1 to ﬁx analytic subspaces of separable F-spaces. The ﬁrst version of Corollary 6.4 below assumes the CH. We present also a sketch of another proof of Corollary 6.4 without the CH and heavily dependent on Mycielski’s theorem about independent functions.

~~Corollary 6.4 Let E := (E, ξ) be a separable F-space. Let F be an analytic subspace of E. Then, under the CH, the codimension of F in E is either ﬁnite or ℵ equals 2 0 .~~

Proof Let G be an algebraic complement of F in E. Assume that the dimension of G is countable. Since G is a countable union of ﬁnite-dimensional subspaces Gn and each Gn (as metrizable, complete, and separable) is analytic, G is analytic. Therefore F and G are analytic, and E endowed with the direct sum topology τ := ξ|F ⊕ξ|G (stronger than ξ) is analytic. Theorem 5.1 is applied to show ξ = τ . Hence F is closed in E. Corollary 2.3 yields that the codimension of F is ﬁnite. 156 6 Weakly Analytic Spaces

For another proof without the CH (see [127]), assume that F has inﬁnite codimension in E; we may assume that F is dense in E. The proof will be completed if we show that there exists a Cantor set D ⊂ E such that D is linearly independent ∩ ={ } and F span D 0 . It is enough to show that for distinct points x1,...,xn of D ∈ ∈ Kn \{ } ∈ N one has 1≤i≤n aixi / F for each tuple (a1,...,an) 0 . For each n ,set n Rn := (x1,...,xn) ∈ E :∃(a1,...,an) = 0 aixi ∈ F . 1≤i≤n

n The proof will be completed if we show that each Rn is of first category in E . Indeed, then Mycielski’s theorem [306] is applied to obtain a Cantor set satisfying n the condition above. Observe that Rn is an analytic subset of E for each n ∈ N: Define f : (Kn \{0}) × En → E := by f ((a1,...,an), (x1,x2,...,xn)) 1≤i≤n aixi . Then f is continuous. Since = −1 −1 Kn \ Rn prEn f (F ), it is enough to note that f (F ) is analytic. As the spaces {0}), E and F are analytic, f −1(F ) is analytic, too. Then [261, S 39.II, Corollary n 1] yields that Rn has the Baire property in E for each n ∈ N. Finally, note that n each Rn is of first category in E . The proof follows by induction: R1 = F is of first category by Theorem 7.2 and the density of F in E. Assume that Rn is of first category in En for some n ∈ N. Set

n+1 A ={(x1,...,xn+1) ∈ E : (x1,...,xn) ∈ Rn} and n+1 B ={(x1,...,xn+1) ∈ E : xn+1 ∈ F + span{x1,...,xn}}. Then A ∪ B equals n+1 (x1,...,xn+1) ∈ E :∃(a1,...,an+1) = 0, aixi + an+1xn+1 ∈ F , 1≤i≤n and the last set is Rn+1. Observe that A = Rn × E. Since Rn is of ﬁrst category in En, it follows that A is of ﬁrst category in En+1. Note also that B ⊂ En × E, and n for each tuple (x1,...,xn) ∈ E the vertical section of B equals

~~B(x1,...,xn) = F + span{x1,...,xn}⊂E.~~

n Clearly, B(x1,...,xn) is analytic and of first category in E. Since E is of first category in En+1 and F is of first category in E, it follows that B has the Baire property in En × E = En+1. Now the fact that B is of first category in En+1 follows from the Kuratowski–Ulam theorem [261]. Clearly, every countable-dimensional tvs is analytic. Proposition 6.7 is taken from [133], although it was already presented in [127], [335] and [378]; see also Young’s theorem in [261], or [181]. 6.4 Trans-separable topological spaces 157

~~Proposition 6.7 If E is an uncountable-dimensional analytic tvs, then the dimen- ℵ sion of E is 2 0 .~~

Proof By the assumption, there exists a continuous map T from NN onto E.Itis enough to show that NN contains a set C homeomorphic to the Cantor set such that N ℵ T(C) is linearly independent. Clearly, dim E ≤ card E ≤ card N = 2 0 .Nowwe prove the reverse inequality. Let G be the family of open subsets G of NN such that

~~ℵ dim T(G):= dim span T(G)≤ 2 0 .~~

N Since N is hereditarily Lindelöf, there exists a countable subfamily S of G such ℵ that S := S = G. Note that dim T(S)≤ 2 0 . Hence, for the closed set D := NN \ S, we note the following two conditions: ℵ (i) If H is a nonempty open set in D, then dim T(H)>2 0 . (ii) If H1,...,Hk are nonempty open sets in D, there exist points d1 ∈ H1,...,dk ∈ Hk such that the points T(d1),...,T(dk) are linearly independent. Moreover, if x1,...,xk are linearly independent points in E, there exist neigh- ∈ borhoods U1,...,Uk of x1,...,xk, respectively, such that each sequence (yi) 1≤i≤k Ui is linearly independent. n For each n ∈ N,setWn := {0, 1} . By induction, for each n ∈ N and s ∈ Wn there exist a point ps ∈ D, a neighborhood Us of xs := T(ps) in E and a closed ball

~~Ks := Ks(ps,rs) ⊂ D~~

~~−1 ⊂ with the center at ps and radius rs~~

~~6.4 Trans-separable topological spaces~~

It turns out that every tvs that admits a resolution consisting of precompact sets is necessarily trans-separable. This fact, due to N. Robertson [342], will be used in the next section. In this section, we collect a couple of results, mostly from [154] and [342] about uniform trans-separable and trans-separable topological vector spaces. A uniform space X is called trans-separable [210], [199] if every uniform cover of X has a countable subcover. Separable uniform spaces and Lindelöf uniform spaces are trans-separable; the converse is not true in general, although every trans-separable pseudometric space is separable. Clearly, a uniform space is trans- separable if and only if it is uniformly isomorphic to a subspace of a uniform product of separable pseudometric spaces. This implies that every uniform quasi-Suslin 158 6 Weakly Analytic Spaces space (see [421, 1.4.2]) is trans-separable. Trans-separable spaces enjoy good per- manence properties; for example, the class of trans-separable spaces is hereditary, productive and closed under uniform continuous images; see [328]. AtvsE is trans-separable if and only if E is isomorphic to a subspace of a product of metrizable separable tvs’s. Thus, if E is an lcs, then (E,σ(E,E)) is trans-separable. A tvs E is trans-separable if and only if for every neighborhood of zero U in E there exists a countable subset N of E such that E = N + U; see, for example, [197], [259], [329], [342]. It is easy to see also that a tvs E is trans- separable if and only if for each continuous F-seminorm p on E the F-seminormed space (E, p) is separable or the associated F-normed space E/ker p is separable. The concept of trans-separability has been used to study several problems, both from analysis and topology; for example, while studying the metrizability of precompact sets in uniform spaces. We refer the reader to papers [82], [342], [153], [369], [154], [131], [241]. Pﬁster [329] observed the following.

~~Proposition 6.8 An lcs E is trans-separable if and only if for every neighborhood of zero U in E its polar U ◦is σ(E,E)-metrizable.~~

~~This fact has been used by Pﬁster [329] to show that precompact sets in (DF )- spaces are metrizable. We will provide below many general results of this type. We start with the following.~~

~~Proposition 6.9 A completely regular topological Hausdorff space X is realcompact if and only if there exists an admissible uniformity N on X such that (X, N ) is trans-separable and complete.~~

Proof If X is realcompact, it is homeomorphic to a closed subset of RC(X). Then the induced uniformity in X is admissible, complete and trans-separable. Conversely, if N is a trans-separable and complete admissible uniformity on X, then (X, N ) is isomorphic to a closed subspace of a product of metrizable separable uniform spaces. Therefore X is realcompact.

~~Corollary 6.5 A completely regular Hausdorff space X is K-analytic if and only if there exists on X an admissible uniformity N such that (X, N ) is complete and admits a compact resolution.~~

Proof If X is K-analytic, then it is realcompact, and Proposition 6.9 applies. Con- versely, by the assumption on X, we know that it is trans-separable. By Proposition 6.9, the space X is realcompact, and by Proposition 3.13 X is K-analytic.

~~Next, Proposition 6.10 is due to N. Robertson [342].~~

~~Proposition 6.10 Let X be a tvs with a precompact resolution. Then X is trans- separable. 6.4 Trans-separable topological spaces 159~~

N Proof Let {Bα : α ∈ N } be a precompact resolution of E. For each α,letKα be = { : ∈ NN} the closure of Bα in the completion Y of X. Set Z α Kα. Then Kα α is a compact resolution in Z. By Proposition 6.2, each metrizable space Z/ker p is separable, where p is a continuous F-seminorm on Z.

~~Corollary 6.6 (Dieudonné) A metrizable tvs E is separable and complete if every bounded set in E is relatively compact. Hence every Fréchet–Montel space is separable.~~

Proof Let (Un)n be a decreasing basis of neighborhoods of zero in E. For each = ∈ NN := { : ∈ NN} α (nk) ,setAα k nkUk. Then Aα α is a bounded resolution on E swallowing bounded sets. Since, by the assumption, every bounded set in E is relatively compact, E has a compact resolution swallowing compact sets. Then E is separable by Proposition 6.10 and is complete by Theorem 6.1.

~~Corollary 6.7 supplements Corollary 3.6.~~

~~Corollary 6.7 For a complete tvs E, the following conditions are equivalent: (i) E has a compact resolution. (ii) E is quasi-Suslin. (iii) E is K-analytic.~~

Proof Only (ii) ⇒ (iii) needs a proof: Since E is trans-separable, as a closed subset of a product of separable and metrizable tvs’s, it is realcompact. Then E is K- analytic by Proposition 3.13.

~~X Since Cp(X) is a (dense) subspace of the product R , the space Cp(X) is always trans-separable. For spaces Cc(X), we have the following simple lemma [154].~~

~~Lemma 6.5 The compact sets in a completely regular Hausdorff space X are metrizable if and only if Cc(X) is trans-separable.~~

Proof If Cc(X) is trans-separable and K is a compact subset of X, then Cc(K) is a separable Banach space since the restriction map T : Cc(X) → Cc(K), T (f ) := f |K, is a continuous surjection. Hence K is a metrizable space. For the converse, assume that {Ki : i ∈ I} is the family of all compact subsets of X, and assume that : → = all Ki are metrizable. Since the map ϕ Cc(X) i∈I Cc(Ki), where ϕ(f) { | : ∈ } f Ki i I is an isomorphism onto its range, and each Cc(Ki) is separable, the conclusion holds.

~~The following result, originally proved in [82], can be shown by using the concept of trans-separability. The proof follows from [299].~~

Proposition 6.11 Let (X, U ) be a uniform space whose uniformity admits a U - N basis B ={Nα : α ∈ N } (i.e., Nα ⊂ Nβ if β ≤ α). Then the precompact subsets of (X, U ) are metrizable in the induced uniformity. 160 6 Weakly Analytic Spaces

Proof By E we denote the lcs of bounded real-valued uniformly continuous functions deﬁned on X endowed with the topology of uniform convergence on precom- N N pact subsets of X. For each α = (m, α1,α2,...,αn,...) ∈ N×(N ) ,byAα denote the set of all f ∈ E such that ≤ | − | ≤ ∈ × ∩ ∈ N f ∞ m, f(s) f(t) 1/n, (s,t) (X X) Nαn ,n . N N Clearly, Aα ⊂ Aβ if α ≤ β and E = {Aα : α ∈ N×(N ) }. Note that each Aα is pointwise compact and, by the equicontinuity, the pointwise topology and the induced topology of E coincide on each Aα . Hence each Aα is a compact subset of E. By Proposition 6.10, E is trans-separable. Thus each equicontinuous subset of the dual E is σ(E,E)-metrizable. Assume that K is a precompact subset of X. Set W := f ∈ E : sup |f(x)| ≤ 1 . x∈K

~~If ϕ is the evaluation map from X into E, the equality ϕ (K) = W ◦ ∩ ϕ (X) yields that K is metrizable.~~

~~The observations above apply to deducing the following theorem [154].~~

~~Theorem 6.3 Compact subsets of an lcs E are metrizable if and only if E endowed with the topology τc of the uniform convergence on compact sets of E is trans- separable.~~

Proof Let F be the dual of (E ,τc).If(E ,τc) is trans-separable, the τc-equicontinuous subsets of (F, σ (F, E)) are metrizable by Proposition 6.8. Since for a com- ◦◦ ◦◦ pact set K in E the bipolar K is τc-equicontinuous, K is metrizable in the topology σ(F,E). Hence K ⊂ E is σ(F,E)-metrizable, and consequently K is σ(E,E)-metrizable. This shows that K is metrizable in E. Conversely, assume that all compact sets in E are metrizable. Cc(E) is trans-separable by Lemma 6.5. Since (E ,τc) is a topological subspace of Cc(E), the conclusion follows.

Since every quasi-Suslin lcs is trans-separable, Theorem 6.3 implies Valdivia’s theorem [421, Theorem 1.4.3 (27)] stating that, if (E ,τc)is quasi-Suslin, all compact sets in E are metrizable. We note the following variant for the topology τp;see [154].

~~Theorem 6.4 Precompact sets are metrizable in an lcs E if and only if E endowed with the topology τp of the uniform convergence on precompact sets of E is trans- separable.~~

Proof If (E ,τp) is trans-separable, by Proposition 6.8 all precompact sets in E are metrizable. To prove the converse, assume that all precompact sets in E are metrizable. Let {Pi : i ∈ I} be the family of all precompact sets in E. For every 6.4 Trans-separable topological spaces 161

i ∈ I ,letKi be the closure of Pi in the completion of E.ThemapT : (E ,τp) → ={ | : ∈ } i∈I Cc(Ki) deﬁned by the form T(u) v Ki i I , where v is the continuous linear extension of u to the completion of E, is an isomorphism onto its range. This yields the conclusion.

Corollary 6.8 The strong dual (E,β(E,E))of an lcs E is trans-separable if and only if every bounded set in E is metrizable in (E, σ (E, E)). Consequently, the strong dual of a (DF )-space E is separable iff every bounded set in E is weakly metrizable.

A family F of functions from a uniform space (X, N ) into a uniform space (Y,M ) is called uniformly equicontinuous [239, Exercise G, Chapter 7] if for each V ∈ M there is U ∈ N such that (f (x) ,f (y)) ∈ V whenever f ∈ F and (x,y) ∈ U. A uniform space (X, N ) is uniformly isomorphic to a subspace of the uniform product of the pseudometric spaces {(X, d) : d ∈ P} endowed with their corresponding pseudometric uniformities Nd . Recall that a topological space X has a countable tightness if for each A ⊂ X and each x ∈ A there exists a countable set B ⊂ A such that x ∈ B. In order to prove Theorem 6.5 below, we need the following lemma from [154].

Lemma 6.6 For a pseudometric space (X, d), the following assertions are equivalent: (i) Every pointwise bounded, uniformly equicontinuous set of functions on (X, Nd ) is a metrizable relatively compact subset of Cc (X, d). (ii) Every pointwise bounded, uniformly equicontinuous set of functions on (X, Nd ) has countable tightness in Cc (X, d). (iii) X is separable.

~~Proof (ii) ⇒ (iii): We may assume that the pseudometric space (X, d) is bounded. Let K (X) be the family of all compact subsets of X. Deﬁne fA (x) := d (x,A) for x ∈ X and A ∈ K (X). Since~~

|fA (x) − fA (y)| ≤ d (x,y) for every A ∈ K (X) and (x,y) ∈ X × X, the family H := {fA : A ∈ K (X)} of real-valued functions on (X, Nd ) is uniformly equicontinuous. This implies that RX H ⊆ C (X, d) . Since (X, d) is bounded, one gets λ>0 with d (x,y) ≤ λ for every (x,y) ∈ X ×X. Hence 0 ≤ fA (x) ≤ λ for each A ∈ K (X) and x ∈ X. There- RX fore the family H is a pointwise bounded, uniformly equicontinuous family of functions on (X, d). Let 0 be the null function on X. Then

~~C (X,d) RX 0 ∈ H c = H .~~

~~By the assumption, there is a sequence {An : n ∈ N} in K (X) such that~~

~~ C (X,d) ∈ : ∈ N c 0 fAn n . 162 6 Weakly Analytic Spaces~~

∈ ∈ N Hence, if x X and ε>0, there is k such that fAk (x) <ε. This proves that := ∞ Y n=1 An is a dense subspace of (X, d). Finally, since every compact pseudometric space An is separable, it follows that (X, d) is separable. (iii) ⇒ (i): Let D be a countable and dense subset of X, and let Z be a closed, pointwise bounded, uniformly equicontinuous subset of Cc (X). Applying Ascoli’s theorem, we note that Z is a compact subspace of Cc (X). Set δx (f ) := f (x) for { | } all f ∈ C (X) and for each x ∈ X. Then δx Z : x ∈ D is a countable family of continuous functions that separates the points of Z. Hence Z, as a subspace of Cc(X), is metrizable.

~~Theorem 6.5 is related to Lemma 6.5.~~

~~Theorem 6.5 A uniform space (X, N ) is trans-separable if and only if every pointwise bounded, uniformly equicontinuous subset of Cc(X, τN ) is metrizable.~~

Proof Assume that every pointwise bounded, uniformly equicontinuous subset of Cc (X) is metrizable. Denote by P the family of all pseudometrics for X generating N . Since the uniformity Nd on X generated by a pseudometric d ∈ P is smaller than N , the assumption above is applied to show that a pointwise bounded, uniformly equicontinuous set A of functions on (X, Nd ) has countable tightness in Cc (X, d). Now Lemma 6.6 shows that the space (X, d) is separable. Since (X, N ) N is uniformly isomorphic to a subspace of the uniform product d∈P (X, d ) of the pseudometric spaces {(X, d) : d ∈ P} endowed with their corresponding pseudometric uniformities, (X, N ) is trans-separable. To prove the converse, assume that (X, N ) is trans-separable. Let C be a closed, pointwise bounded, uniformly equicontinuous subset of Cc (X, τN ). Since | | ∞ ∈ supf ∈C f (x) < for each x X,themap

→ | − | (x,y) supf ∈C f (x) f (y) deﬁnes a pseudometric D (x,y) on X. Then C is a pointwise bounded, uniformly equicontinuous subset of C (X, D). On the other hand, for any ε>0 there is a vicinity U in X × X such that | − | supf ∈C f (x) f (y) <ε for every (x,y) ∈ U. Hence D (x,y) <ε for every (x,y) ∈ U, so the identity map from (X, N ) onto (X, ND) is uniformly continuous. Since (X, N ) is trans- separable, (X, D) is separable, and Lemma 6.6 is applied to show that C is a compact metrizable subspace of Cc (X, D). Finally, since the map

T : Cc (X, D) → Cc (X, τN ) deﬁned by Tf = f ◦ ϕ, where ϕ is the identity from (X, τN ) onto (X, D),isin- jective and continuous, it follows that C is also a metrizable compact set when considered as a subspace of Cc (X, τN ). The proof is complete. 6.4 Trans-separable topological spaces 163

By Ascoli’s theorem, every closed, pointwise bounded, uniformly equicontinuous set in Cc(X) is compact. Hence we have the following corollary. Corollary 6.9 Let (X, N ) be a uniform space. If every compact subset of Cc (X, τN ) has countable tightness, then (X, N ) is trans-separable. Proposition 6.12 shows that every web-compact space is trans-separable.

~~Proposition 6.12 Let (X, N ) be a uniform space. If the space Cc (X, τN ) is angelic, (X, N ) is trans-separable. In particular, a uniform space X with a precompact resolution is trans-separable.~~

Proof If Cc (X, τN ) is angelic, each compact subset of Cc (X, τN ) has countable tightness. By Corollary 6.9, the uniform space (X, N ) is trans-separable. Let {Kα : α ∈ NN} be a precompact resolution in (X, N ). Since the space Y covered by the closure of the sets Kα in the uniform completion of X is web-compact, we apply Theorem 4.5, and next we apply the ﬁrst part of the proposition already proved.

~~Corollary 6.10 Let X be a metric space. Then Cp (X) is angelic if and only if Cc (X) is angelic if and only if X is separable. We show that the converse in Corollary 6.9 fails.~~

~~Example 6.2 There exists a trans-separable space (X, N ) such that the space Cc (X, τN ) contains a compact set K that does not have countable tightness.~~

Proof Let ζ be the first ordinal of an uncountable cardinality, and let ϕζ denote the locally convex direct sum of |ζ | > ℵ0 copies of R. Then the dual of ϕζ is isomorphic [0,ζ ) to the product ωζ = R of |ζ | copies R. If X denotes the linear space ϕζ with the weak topology σ(ϕζ ,ωζ ), then X is a uniform space under the associated uniformity N . Hence τN = σ(ϕζ ,ωζ ) and X = ωζ . Then (X, N ) is trans-separable. Since the topology of uniform conver- [0,ζ ) gence on weakly compact subsets of ϕζ and the product topology of R coincide on ωζ , we conclude that ωζ is isomorphic to a subspace of Cc (X, τN ). Consider [0,ζ ) the compact set K := [0, 1] of ωζ , and let f : [0,ζ) → R be the constant function such that f (γ ) = 1 for each γ ∈ [0,ζ). Then, if F is a finite subset of [0,ζ), define gF : [0,ζ) → R so that gF (γ ) = 1ifγ ∈ F and gF (γ ) = 0ifγ ∈ [0,ζ) \ F. Let F (ζ) be the family of all finite subsets of [0,ζ). Set

~~A := {gF : F ∈ F (ζ )} .~~

ωζ Then f ∈ A ⊆ K, and no countable set {Fn : n ∈ N}⊆F (ζ) veriﬁes f ∈ ω : ∈ N ζ ∈ \ ∞ gFn n . Indeed, if ξ [0,ζ) n=1 Fn, then 1 U (f,ξ) := h ∈ R[0,ζ ) : |h (ξ) − 1| < 2 ∈ ∈ ∈ N is a neighborhood of f ωζ such that gFn / U (f,ξ) for every n . 164 6 Weakly Analytic Spaces

~~6.5 Weakly analytic spaces need not be analytic~~

Let E be an lcs. By μ(E, E) and β(E,E) we denote the Mackey and strong topologies of E, respectively. By the Mackey and strong duals of E we understand E endowed with the Mackey topologies μ(E,E) and β(E,E), respectively. The topology μ(E,E) is the strongest locally convex topology on E compatible with the dual pair (E,E);see[213]or[240] for details. We have already noticed (see Theorem 12.8) that the weak topology σ(E,E) of a WCG Banach space E is K-analytic, and there exist nonseparable WCG Banach spaces. The following problem seems to be interesting. When can analyticity or K- analyticity of the weak topology σ(E,E) of a dual pair (E, E) be lifted to stronger topologies on E compatible with the dual pair? Note that if (E, σ (E, E)) is analytic, E endowed with a stronger topology ξ on E is analytic if and only if ξ admits a (relatively countably) compact resolution. Indeed, the analyticity of σ(E,E) yields a weaker metric topology on E by Propo- sition 6.3. Hence ξ is also angelic, and Corollary 3.6 completes the proof. There exist K-analytic spaces Cp(X) such that the space Cp(Cp(X)) is not even a Lin- delöf space; see [24, Example 7.14]. This follows from the following fact due to Reznichenko.

~~Example 6.3 There exists a compact space X and x ∈ X such that X = β(X \{x}) and Cp(X) is K-analytic.~~

Indeed, since Y := X \{x} is pseudocompact, not compact, Y is not realcompact. Hence Cp(Cp(Y )) is not realcompact, so it is not Lindelöf. On the other hand, Cp(Y ), as a continuous image of the K-analytic space Cp(X), is a K-analytic space. We will use the following proposition; for the proof we refer the reader to [27, Corollary 0.5.14, Proposition 0.5.12].

∗ Proposition 6.13 Let Lp(X) be the weak dual of Cp(X). Then X is a Lindelöf Σ-space (K-analytic space, analytic space, separable, or σ -compact) if and only if Lp(X) is a Lindelöf Σ-space (K-analytic space, analytic space, separable, σ - compact).

Ferrando [151] proved that the space L[0, 1] endowed with the Mackey topology μ(L[0, 1],C[0, 1]) is a weakly analytic space that is not K-analytic. This result has been extended in [236] to the present form.

~~Theorem 6.6 (Kakol–López-Pellicer– ˛ Sliwa)´ For a completely regular Hausdorff space X, the Mackey dual of Cp(X) is analytic if and only if X is countable.~~

Proof Set X := (X, τ), where τ is the original topology of X.ByLμ(X) denote the Mackey dual of Cp(X) (i.e., the dual of Cp(X) endowed with the Mackey topology μ = μ(Cp(X) ,Cp(X))). Assume that Lμ(X) is analytic. Suppose, by a contradiction, that X is uncountable. For x ∈ X the functional δx : Cp(X) → R deﬁned by 6.5 Weakly analytic spaces need not be analytic 165

δx(f ) = f(x) is linear and continuous. Denote by Lp(X) the dual of Cp(X) en- ∗ dowed with the weak dual topology σ = σ(Cp(X) ,Cp(X)). Set Y ={δx : x ∈ X}. The map δ : (X, τ) → (Y, σ |Y) deﬁned by x → δx is a homeomorphism, and the set Y is closed in Lp(X);see [27, Proposition 0.5.9]. Hence Y is also closed in Lμ(X). Thus (Y, μ|Y)is analytic. Let γ be the topology on X such that δ is a homeomorphism between (X, γ ) and (Y, μ|Y). Since (X, γ ) is an uncountable analytic space, it contains a set A homeomorphic to the Cantor set; see, for example, [346]. Clearly, γ |A = τ|A. Let (xn)n ⊂ A be a sequence such that xn = xm for n = m that converges to some

~~x0 ∈ (A \{xn : n ∈ N}).~~

~~It is easy to see that for every closed subspace G of (X, τ) and every x ∈ (X \ G) there exists f ∈ C(X,I) with f(x)= 1 such that G ∩ suppf =∅. Put~~

Xn ={xk : k>n}∪{x0} for n ∈ N. Clearly, Xn is closed in X and xn ∈ Xn for n ∈ N. Therefore, we can construct inductively a sequence (fn)n ⊂ C(X,I) such that fn(xn) = 1 and n−1 supp fn ∩ Xn ∪ supp fk =∅. k=1 ∈ Then x0 k supp fk and

supp fn ∩ supp fm =∅ for all n, m ∈ N with n = m. Denote by Cb(X) the Banach space of all bounded real-valued continuous functions on X endowed with the sup norm ·.Letg ∈ Cb(X) .Fork ∈ N,set

αk =|g(fk)|/g(fk) if g(fk) = 0 and set αk = 1 otherwise. Then |αk|=1 and αkg(fk) =|g(fk)| for k ∈ N. ∈ N = n ∈ = Let n and Sn k=1 αkfk. Then Sn Cb(X) and Sn 1. Thus n n | |= =| |≤ g(fk) αkg(fk) g(Sn) g k=1 k=1 for n ∈ N,so ∞ |g(fk)|≤g. k=1 166 6 Weakly Analytic Spaces

~~Hence g(fk) → 0. It follows that the sequence (fn)n converges weakly to 0 in Cb(X). Thus the set~~

F0 ={0,f1, −f1,f2, −f2,...} is weakly compact in Cb(X). By Krein’s weak compactness theorem [287, Theo- rem 2.8.14], the closed convex hull F of F0 in Cb(X) is weakly compact. Clearly, F is the closed absolutely convex hull of the set {fk : k ∈ N} in Cb(X). The topology of the pointwise convergence in Cb(X) is weaker than the weak topology of Cb(X), so F is compact in (Cb(X), ). Hence F is compact in Cp(X) since the injection map (Cb(X), ) → Cp(X) is continuous. Thus the functional

~~pF : Lμ(X) →[0, ∞), deﬁned by~~

pF (g) = sup{|g(f )|:f ∈ F }, is a continuous seminorm. Since (fn)n ⊂ F, we have ≥| |= pF (δxn ) fn(xn) 1 ∈ N = ∈ = for n . It is easy to see that f(x0) 0 for all f F ,sopF (δx0 ) 0. It follows → | → that δxn δx0 in (Y, μ Y),soxn x0 in (X, γ ), a contradiction. Now assume that X is countable. If Cp(X) is finite-dimensional, the Mackey dual Lμ(X) of Cp(X) is finite-dimensional, so it is analytic. If Cp(X) is infinite- N dimensional, Cp(X) is a metrizable lcs isomorphic to a dense subspace of R . N Hence Lμ(X) is algebraically isomorphic to ϕ, the strong dual of R . Since ϕ endowed with the strongest locally convex topology is the union of an increasing sequence of finite-dimensional Banach spaces, it is an analytic space. It follows that Lμ(X) is analytic.

~~Theorem 6.6 and its proof also yield the following corollary.~~

~~Corollary 6.11 The strong dual of Cp(X) is analytic if and only if X is countable.~~

~~Recall that Lp(X) is analytic if and only if X is analytic by Proposition 6.13. Thus Theorem 6.6 provides many concrete nonanalytic lcs’s whose weak topology is analytic.~~

~~Corollary 6.12 Let X be an uncountable analytic space. Then the Mackey dual Lμ(X) of Cp(X) is weakly analytic but not analytic.~~

Note that in Corollary 6.12 the Mackey dual Lμ(X) is not even K-analytic. Indeed, ∗ since Lμ(X) is weakly analytic, the weak dual of Cp(X) admits a weaker metric topology by Proposition 6.3. Assume that Lμ(X) is K-analytic. Then, by the ﬁrst part of Proposition 6.3,wehavethatLμ(X) is analytic, a contradiction with Corollary 6.12. 6.6 More about analytic locally convex spaces 167

~~6.6 More about analytic locally convex spaces~~

Corollary 6.12 might be a good motivation to study sufﬁcient conditions for a weakly analytic lcs to be analytic in the original topology. Let E be a vector space, and let ξ ≤ τ be two vector topologies on E. Recall that τ is called ξ-polar if it admits a basis of ξ-closed neighborhoods of zero. It is easy to see that if A is a subset of E that is complete (sequentially complete) in ξ,it is complete (sequentially complete) in τ ;see[213, Theorem 3.2.4]. This yields the following corollary.

~~Corollary 6.13 If τ is ξ-polar, a ξ-complete (sequentially complete) resolution on E is τ -complete (sequentially complete).~~

~~This combined with Proposition 6.3 and Corollary 6.2 provides the following.~~

~~Corollary 6.14 If τ is a metrizable and separable vector topology on a vector space E such that ξ ≤ τ , τ is ξ-polar, and (E, ξ) has a complete resolution, then (E, τ) is analytic.~~

Since for every separable lcs E its weak∗ dual topology σ(E,E) admits a weaker metric topology, we have that for separable E any topology ξ stronger than σ(E,E) is analytic if ξ admits a compact resolution. It turns out that the following general fact holds [236].

Theorem 6.7 A separable tvs E := (E, ξ) having a sequentially complete resolution is analytic if E satisﬁes one of the following conditions: (i) E is covered by a sequence (Sn)n of absolutely convex metrizable subsets. (ii) E is a continuous linear image of a separable and metrizable tvs.

N Proof Assume that the family {Aα : α ∈ N } is a sequentially complete resolution in the space E. Assume (i). Then, by [213, Theorem 9.2.4], it follows that each (Sn,ξ|Sn) is N metrizable, and then it has the complete resolution {Sn ∩ Aα : α ∈ N }. Since ξ is separable, (E, ξ) is trans-separable. Then each metrizable and trans-separable subspace (Sn,ξ|Sn) is separable. By Corollary 6.14, each space (Sn,ξ|Sn) is analytic, so (E, ξ) is analytic. Assume (ii). First note that (E, ξ) admits a stronger separable and metrizable vector topology τ .Letτ ξ be a vector topology on E whose neighborhoods of zero are composed by the ξ-closures of τ -neighborhoods of zero. Then ξ ≤ τ ξ ≤ τ , τ ξ is metrizable, separable and ξ-polar. The space (E, τ ξ ) admits a complete resolution. Hence, applying Corollary 6.14, (E, τ ξ ) and (E, ξ) are analytic.

~~Corollary 6.15 Every Montel (DF )-space is analytic.~~

~~Proof Let (Sn)n be a fundamental sequence of bounded absolutely convex closed subsets of E. The strong dual (E,β(E,E))is a Montel Fréchet space, so the space 168 6 Weakly Analytic Spaces~~

~~(E,β(E,E)) is separable. Then every bounded set in (E,σ(E,E)) is metrizable. Hence every bounded closed set in E is metrizable and compact, and Theorem 6.7 applies.~~

~~Theorem 5.1 and Corollary 6.14 supplement Proposition 3.14 by noting the following corollary.~~

~~Corollary 6.16 Let E be a separable Fréchet space. Then no proper dense, ﬁnite- codimensional subspace F of E is weakly K-analytic.~~

Proof Assume F is weakly K-analytic. By Corollary 6.13, the space F has a complete resolution in the relative topology of E. Applying Corollary 6.14, we deduce that F is analytic. Let D be a ﬁnite-dimensional algebraic complement of F to E. Let q : E → E/D be the quotient map. Since F is a proper dense subspace of E, q|F : F → E/D is an injective continuous map that is not an isomorphism. There- fore on F there exists a strictly weaker metrizable and complete locally convex topology ξ such that (F, ξ) is isomorphic to E/D. Since F is analytic, we reach a contradiction by applying Theorem 5.1 for the identity map (F, ξ) → F .

~~6.7 Weakly compact density condition~~

If E is a Banach space, the Mackey dual (E,μ(E,E)) is not metrizable, except when E is reﬂexive. It is well known that (E,μ(E,E)) is a complete lcs; see, for example, [246]. On the other hand, if B is the unit ball in the dual E of E, one may expect that under some conditions (B,μ(E,E)|B) will be metrizable. Schlüchtermann and Wheller [367] introduced the class of strongly weakly compactly generated (SWCG) Banach spaces. A Banach space is SWCG if the space (B,μ(E,E)|B) is metrizable; see also [368]. Theorem 6.8, from [367, Theorem 2.1], shows that every SWCG Banach space is WCG. In [367, Theorem 2.5], it is proved that every SWCG Banach space is weakly sequentially complete. Hence the space c0 being a WCG space is not SWCG.

Theorem 6.8 Let E be a Banach space. Let B and B be the closed unit balls in E and E, respectively. The following conditions are equivalent: (i) (B,μ(E,E)|B) is metrizable. (ii) There exists a sequence (Kn)n of weakly compact, absolutely convex subsets of E such that for every weakly compact set L ⊂ E and every ε>0 there exists n ∈ N such that L ⊂ Kn + εB. (iii) There exists a weakly compact, absolutely convex set K ⊂ E such that for each weakly compact set L ⊂ E and every ε>0 there is n ∈ N such that L ⊂ nK + εB. 6.7 Weakly compact density condition 169

Proof (i) ⇒ (ii): Since (B,μ(E,E)|B) is metrizable, there exists a sequence ◦ ∩ (Kn)n of weakly compact absolutely convex sets in E such that (Kn B )n is a countable basis of neighborhoods of zero in μ(E,E)|B.LetL be a weakly compact set in E, and choose ε>0. We may assume that ε<1. Set c = ε−1.Fixn ∈ N such that ◦ ∩ ⊂ ◦ ∩ Kn B (cL) B . Then ◦ ∩ ◦ ⊂ ◦ ∩ ◦ ((cL) B ) (Kn B ) . This implies that ⊂ ◦◦ ∪ ◦◦ ⊂ ◦◦ ∪ ◦◦ ⊂ + cL ((cL) B) (Kn B) Kn B, and this yields L ⊂ Kn + εB. (ii) ⇒ (iii): Let (Kn)n be a sequence as assumed in (ii). Let

tn := sup{x:x ∈ Kn} for each n ∈ N. Set := −n −1 : ∈ K 2 tn xn xn Kn . n Note that K is an absolutely convex weakly closed set in E. It is easy to see that K is weakly compact in E, and we can also use Grothendieck’s characterization of the n weak compactness; see [123, p. 227]. Note that Kn ⊂ 2 tnK. (iii) ⇒ (i): Deﬁne a metric

d(f,g) := max |f(x)− g(x)| x∈K for f,g ∈ E. Clearly, the topology ξ generated by this metric is weaker than μ(E,E). We prove that μ(E,E)|B ≤ ξ|B. Let (ft )t∈T be a net in (B ,ξ|B ) that converges to f .IfL is a weakly compact set in E and ε>0, there exists n ∈ N such that

~~− L ⊂ nK + ε(4) 1B.~~

~~There exists t0 ∈ T such that for t>t0 one has~~

−1 |ft (y) − f(y)| <ε(2n) for all y ∈ K. Then |ft (x) − f(x)| <εfor all t>t0 if x ∈ L because then there −1 exists y ∈ K such that x − ny≤ε(4) . This proves that (ft )t∈T converges to f in the topology μ(E,E). 170 6 Weakly Analytic Spaces

If E is a separable SWCG Banach space, (E ,μ(E ,E))is separable. Since E = n nB and each nB is metrizable, by Theorem 6.7 the space (E ,μ(E ,E)) is analytic. Therefore we have the following proposition.

~~Proposition 6.14 Let E be an SWCG Banach space. Then (E,μ(E,E)) is analytic if and only if E is separable.~~

Let E be a Banach space and let (S,Σ,μ) be a finite measure space. By L1(μ, E) we denote a Banach space of the Bochner integrable functions f : S → E. Schlüchtermann and Wheller [367, Theorem 3.2] asked when L1(μ, E) is SWCG. Talagrand [394](seealsoDiestel[122]) proved that L1(μ, E) is WCG if E is a WCG Banach space. If E is a separable Banach space, the Mackey dual (E,μ(E,E)) is separable but (E,β(E,E)) need not be separable. Clearly, (E,β(E,E)) is analytic if and only if (E,β(E,E)) is separable. Theorem 6.8 and Proposition 6.14 may suggest the following question: Let E be a separable Banach space. Is it true that the Mackey dual (E,μ(E,E))of E is an analytic space? In this section, we provide some sufficient conditions for a separable (LF )-space E to have the Mackey dual (E,μ(E,E))analytic. Now we recall the concept of the density condition (dc) introduced by Heinrich in [203]; see also [55]. Let E be a metrizable lcs with a countable basis (Un)n of absolutely convex neighborhoods of zero. We will say that E satisfies the density condition if there exists a double sequence (Bn,k)n,k of bounded sets in E such that for each n ∈ N and each bounded set C ⊂ E there is k ∈ N such that C ⊂ Bn,k + Un. For Proposition 6.15, we refer to Bierstedt, and Bonet’s work [55].

~~Proposition 6.15 Every Fréchet–Montel space (i.e., a Fréchet space for which every bounded closed set is compact) satisﬁes the density condition.~~

Proof Let (Un)n be a countable basis of neighborhoods of zero in E such that Un+1 + Un+1 ⊂ Un for each n ∈ N. Since E is Fréchet–Montel, E is separable by Corollary 6.6.Let{xk : k ∈ N} be a countable dense subset of E. For each n, k ∈ N, let Bn,k be the absolutely convex envelope of the ﬁnite set {x1,x2,...,xk}.Fixn ∈ N and a bounded set C ⊂ E. Then (since C is precompact) there exists a ﬁnite set F ⊂ E such that C ⊂ F + Un+1. As the set {xk : k ∈ N} is dense in E, there exists k ∈ N such that F ⊂ Bn,k + Un+1. Then

~~C ⊂ Bn,k + Un+1 + Un+1 ⊂ Bn,k + Un.~~

~~The proof is completed.~~

Note that there exist reﬂexive Fréchet spaces whose strong dual is separable (such spaces are distinguished, see Section 16.1) and that do not satisfy the density condition [51]. The next proposition was obtained by Bierstedt and Bonet in [51]. 6.7 Weakly compact density condition 171

~~Proposition 6.16 A metrizable lcs E satisﬁes the density condition if and only if every bounded set in (E,β(E,E))is metrizable.~~

Proof Let (Un)n be a decreasing basis of absolutely convex neighborhoods of zero for E such that Un+1 + Un+1 ⊂ Un for each n ∈ N. It is known that the bounded ◦ sets in (E ,β(E ,E)) are metrizable if and only if each polar Un has a countable basis of neighborhoods of zero in the topology β(E,E);see[213, Lemma 9.2.4]. Therefore, we need only to show that for each bounded absolutely convex closed set C ⊂ E, and for each n ∈ N, there exists a bounded set Bn,k ⊂ E such that

~~◦ ∩ ◦ ⊂ ◦ Bn,k Un+1 C .~~

~~The rest of the proof follows easily from the bipolar theorem [213, Theorem 8.2.1]:~~

~~⊂ ◦ ∩ ◦ ◦ ⊂ + + ⊂ + C (Bn,k Un+1) Bn,k Un+1 Un+1 Bn,k Un.~~

Then ◦ ∩ ◦ ⊂ + ◦ ⊂ ◦ Bn,k Un 2(Bn,k Un) 2C . Let K(E) be the family of all absolutely convex σ(E,E)-compact sets in an lcs E. Motivated by papers [367], [51], [54], [55], we will say that a metrizable lcs E with a countable basis (Un)n of absolutely convex neigbourhoods of zero in E satisﬁes the weakly compact density condition (wcdc) if there is in K(E) a double sequence (Bn,k)n,k such that for n ∈ N and C ∈ K(E) there exists k ∈ N such that C ⊂ Bn,k + Un. If K(E) denotes the family of all bounded sets in E, the condition above describes the density condition for a metrizable lcs E. For Fréchet–Montel spaces, wcdc and dc are equivalent. We discuss the conditions wcdc and dc for Köthe echelon spaces. Consider the class of Köthe echelon spaces λp := λp(I, A), where A = (an) is any Köthe matrix on a countable set I with 1

The space λ1 satisﬁes the wcdc for any A. Indeed, let {xn : n ∈ N} be a dense countable subset in λ1, and let (Un)n be a countable basis of neighborhoods of zero in λ1. Then, for a weakly compact set C ⊂ λ1 and n ∈ N there exists k ∈ N such that k C ⊂ aj xj :|aj |≤1 + Un. j=1

(ii) The strong dual (λ1,β(λ1,λ1)) is analytic if and only if λ1 is Montel. Indeed, if λ1 is Montel, its strong dual is covered by a sequence of absolutely convex, compact metrizable sets and Theorem 6.7 applies. The converse follows from the Dieudonné–Gomes theorem [288, Theorem 27.9]. (iii) If λ satisﬁes the dc and is not Montel (see [51, Theorem 4, Corollary 8] 1 describing this case), then (λ1,β(λ1,λ1)) is not quasi-Suslin. Indeed, by the condition dc, every closed bounded set in the (DF )-space (λ ,β(λ ,λ1)) is metrizable (and complete by the completeness of β(λ ,λ1)). As- 1 1 1 sume (λ ,β(λ ,λ1)) is quasi-Suslin. Since (λ ,μ(λ ,λ1) is analytic, it admits a 1 1 1 1 weaker metric topology by Proposition 6.3. Hence (λ ,β(λ ,λ1)) is K-analytic. By 1 1 Proposition 6.3, the space (λ1,β(λ1,λ1)) is analytic. Hence λ1 is Montel. (iv) There exists a separable reﬂexive Fréchet space E that does not satisfy the dc and where (E ,β(E ,E))is analytic. Indeed, let A = (an) be a Köthe matrix on N satisfying the condition ND; see [328]. Then λp for p>1 does not satisfy the dc [51, p. 178]. On the other hand, the strong dual of λp is analytic. Proposition 6.17 motivates Theorem 6.9 below.

Proposition 6.17 Let E be a separable (LF )-space. Then the dual E endowed with the topology τpc(E ,E) of the uniform convergence on precompact sets of E is an analytic space. Hence (E,σ(E,E))is also an analytic space.

Proof Let (En)n be a sequence of Fréchet spaces deﬁning the space E. For each ∈ N n n ,let(Uk )k be a decreasing basis of absolutely convex neighborhoods of zero E α = (n ) ∈ NN A := n (U k )◦, (U k )◦ in n.For k ,set α k k nk where nk is the polar in E U k {A : α ∈ NN} E of the set nk . Then α is a resolution in . Since each sequence in any Aα is equicontinuous, each set Aα is relatively compact in the topology τpc(E ,E). By the separability of E, there exists on E a metric topology weaker than the weak topology σ(E,E). Hence (E,σ(E,E)) is angelic, and we apply Theorem 4.1 to deduce that (E ,τpc(E ,E)) is angelic. Next, we apply Corollary 3.6 to get that (E ,τpc(E ,E))is K-analytic. Finally, Proposition 6.3 is used to show that (E ,τpc(E ,E))is analytic.

Having in mind this fact, one can ask about conditions under which the Mackey dual (E,μ(E,E))of a separable (LF )-space E is analytic. The following theorem extends Valdivia’s [421, Theorem 23, p. 77]. We shall need the following lemma due to Valdivia [421, (22), p. 76]. 6.7 Weakly compact density condition 173

Lemma 6.7 Let E be an (LF )-space with a deﬁning sequence (En)n of separable Fréchet spaces. Let A be a compact, absolutely convex set in σ(E,E). Then there exists m ∈ N such that A ⊂ Em and A is bounded in Em.

Proof By EA we denote the linear span of A endowed with the Minkowski func- = ∩ tional Banach topology. Since EA n EA En, by the Baire category theorem there exists m ∈ N such that EA ∩Em is a dense and Baire subspace in the space EA. Clearly, the inclusion

T : EA ∩ Em → (Em,τm) has a closed graph, where τm is the original Fréchet separable topology of Em. By Theorem 5.1,themapT is continuous. To complete the proof, it is enough to observe that EA ⊂ Em.

~~We are ready to prove the following theorem [236].~~

~~Theorem 6.9 Let E be an (LF )-space and (En)n a deﬁning sequence of E of separable reﬂexive Fréchet spaces satisfying the wcdc. Then the Mackey dual (E,μ(E,E))is an analytic space.~~

∈ N := ◦ Proof First assume that E is a metrizable space. Fix n , and set Sn Un , where (Un)n denotes a decreasing basis of absolutely convex closed neighborhoods of zero in E. Applying the wcdc, we deduce that there exists a sequence (Bn,k)k in K(E) with the desired properties. Since the polars of absolutely convex σ(E,E)-compact sets A in E compose a basis of neighborhoods of zero for (E,μ(E,E)), we deduce that for a μ(E,E)-neighborhood of zero V there is k ∈ N such that

◦ ∩ ⊂ Bn,k Sn 2V. | := = This yields the metrizability of (Sn,μ(E ,E) Sn).ThesetsAα Sn1 ,forα ∈ NN (nk) , generate a compact resolution in σ(E,E), so the assumptions of Theo- = rem 6.7 are satisfied. Therefore (E, μ(E ,E)) n Sn is analytic. Now assume that E is an (LF )-space and (En)n is a defining sequence of separable reflexive Fréchet spaces satisfying the wcdc. Since each En is a separable and reflexive space, the strong dual (En,β(En,En)) is analytic (by the previous case), and the projective limit

:= (E ,γ) Projn(En,β(En,En)) is a closed subspace of the analytic space n(En,β(En,En)). As closed subspaces of analytic spaces are analytic, (E,γ) is analytic. Since every absolutely convex σ(E,E )-compact set is contained and bounded in some Em by Lemma 6.7,we deduce μ(E ,E)≤ γ .Ifjn : En → E is the inclusion map, the dual map : → jn (E ,μ(E ,E)) (En,μ(En,En)) 174 6 Weakly Analytic Spaces

~~∈ N = is continuous for n . This, combined with the equality μ(En,En) β(En,En) (since En’s are reﬂexive), yields γ ≤ μ(E ,E).~~

The Mackey dual (E,μ(E,E))of a Banach space has also been studied in [368] and [243]. In [368], the authors proved that if E is a separable SWCG Banach space, the Mackey dual (E ,μ(E,E )), which is clearly analytic, is an ℵ0-space (i.e., it has a countable pseudobase). Recall that a family P of subsets of a topological space E is called a pseudobase if for any open set U ⊂ E and compact set K ⊂ U there exists P ∈ P with K ⊂ P ⊂ U.Everyℵ0-space is separable and Lindelöf [290], [368, Theorem 4.1]. Kirk [243] studied the Mackey dual for Banach spaces C(K) with compact K. On the other hand, from Batt and Hiermeyer [46,2.6](seealso [367], [368, p. 274 and Theorem 4.2]), there exists a separable Banach space E for which (E, σ (E, E )) is not an ℵ0-space. It is known also ([290], [368, Theorem 4.1]) that a regular topological space is both an ℵ0-space and a k-space if and only if it is a quotient of a separable metric space. It seems to be natural to ask about conditions for a Banach space E to have the space (E, σ (E, E)) be a k-space. Recall also that a Hausdorff space X is a k-space if a set A ⊂ X is closed in X if and only if A ∩ K is closed in K for each compact set K ⊂ X. We need the following simple fact due to Grothendieck [191, p. 134].

~~Lemma 6.8 Let E be a Banach space, and let A ⊂ E be a μ(E,E)-compact set in E. Then every σ(E,E)-convergent sequence in E converges uniformly on A.~~

~~Now we are ready to prove the following general proposition.~~

~~Proposition 6.18 If E is a Banach space for which (E, σ (E, E)) is a k-space, the space E is ﬁnite-dimensional.~~

Proof Let γ be the topology on E of the uniform convergence on μ(E,E)-compact sets. Clearly, σ(E,E) ≤ γ . Since σ(E,E) and γ have the same sequentially compact sets by applying Lemma 6.8, both topologies have the same compact sets (note that σ(E,E) and γ are angelic). Assume (E, σ (E, E)) is a k-space. Then σ(E,E ) = γ .Let(xn)n be a null sequence in the norm topology of E . Since {0}∪{xn : n ∈ N} is μ(E ,E)-compact, the sequence (xn)n has a ﬁnite-dimensional linear span. This implies that E (hence also E) is ﬁnite-dimensional.

~~6.8 More examples of nonseparable weakly analytic tvs~~

AtvsE is called dual-separating if its topological dual E separates points of E (i.e., for each x = 0 there exists f ∈ E such that f(x) = 0). The next example provides a nonseparable non locally convex dual-separating F-space E such that the weak topology σ(E,E) is analytic. Example 6.4 uses some arguments from [132]. We need the following simple general fact. If E is a vector space admitting two metrizable vector topologies ξ1 and ξ2 such that ξ1 ≤ ξ2, and every ξ1-bounded set is 6.8 More examples of nonseparable weakly analytic tvs 175

ξ2-bounded, then ξ1 = ξ2. Indeed, take a null sequence (xn)n in ξ1. There exists an unbounded scalar sequence tn ∞such that tnxn → 0inξ1. Hence the sequence = −1 → (tnxn)n is ξ2-bounded, and then xn tn (tnxn) 0inξ2.

Example 6.4 There exists a nonlocally convex, nonseparable, metrizable and com- = plete tvs λ0 (λ0,ξ) such that (λ0,σ(λ0,λ0)) is isomorphic to a (dense) vector subspace of RN. Moreover, (i) (λ0,σ(λ0,λ0)) is analytic, unordered, Baire-like and not Baire, RN \ (ii) (λ0,σ(λ0,λ0)) is a Baire space, and NN (iii) (λ0,σ(λ0,λ0)) contains a copy of as a closed subset.

Proof By [132], there exists a nonlocally convex nonseparable F -space λ0 with a N basis (Un)n of balanced neighborhoods of zero closed in R (i.e., λ0 is the space of all sequences x = (n) of real numbers such that tx→0ast → 0, where := := −1 n { | |} x sup x n, x n n j=1 min 1, j ). Set

−1 Un := {x ∈ λ0 :x≤n } for n ∈ N. The space λ0 endowed with the topology ξ generated by the F-norm . is metrizable, complete and nonseparable, and the topological dual of λ0 is an ℵ0- dimensional vector space; we refer the reader to [132] to check details. We prove the claims (i), (ii) and (iii). ∈ N = (i) Note that for m one has λ0 n nUm, and each nUm is σ(λ0,λ0)- analytic (as a complete metrizable and separable space). Hence the space E := (λ0,σ(λ0,λ0)) is analytic. Since σ(λ0,λ0) is metrizable, the topology σ(λ0,λ0) equals the finest locally convex topology ξc on λ0 weaker than ξ. Indeed, the topologies σ(λ0,λ0) and ξc are metrizable and locally convex, and they generate the same continuous linear functionals. Hence both metrizable topologies have the same bounded sets, so they coincide. The space (λ0,ξc) is an unordered Baire-like space [119]. Indeed, if (An)n is a sequence of absolutely convex closed subsets in the space (λ0,ξc) covering (λ0,ξc), then each set An is closed in the topology ξ. The Baire category theorem is applied to find a number m ∈ N such that Am has a nonvoid interior in ξ. Since Am is an absolutely convex set, using the definition of the topology ξc, we deduce that Am is a ξc-neighborhood of zero. On the other hand, (λ0,ξc) is not Baire. Indeed, otherwise the identity map I : (λ0,ξc) → (λ0,ξ) (which has a closed graph) would be continuous by using the closed graph theorem [1] (between Baire tvs’s and F- spaces), which is impossible (since (λ0,ξ)is not locally convex). N N (ii) Since R \ E is a Gδ-subset of R , Proposition 2.8 is applied so we can conclude that RN \ E is a Polish space and hence Baire. (iii) Note that E cannot be covered by a sequence of bounded sets. Indeed, let N F be the closure of E in R . Clearly, the space F is linearly homeomorphic to the RN = product space . Assume that E n Sn is the union of a sequence of bounded, closed, absolutely convex sets. Since E is an unordered Baire-like space, some Sm is a neighborhood of zero in E. Hence its closure (in the space F ) is a neighborhood 176 6 Weakly Analytic Spaces of zero in F . Consequently, RN is a normed space yielding a contradiction. Hence the space E is not a σ -compact space. Now Theorem 3.12 is applied to get the conclusion of (iii).

For Example 6.5, we need the following deﬁnitions. A function f from [0, 1] into a vector space E is called measurable if all values of f lie in a ﬁnite-dimensional subspace F ⊂ E (depending on f ) and f is Lebesque measurable, where F is endowed with the unique Hausdorff vector topology. The set S(E) of equivalent measurable (classes of) functions is a vector space. Since two constant functions agree almost everywhere if and only if they are identical, there is an injective map tE : E → S(E) that assigns to every x ∈ E the constant function f(t) = x for t ∈[0, 1]. A few additional results will be used to present Example 6.5.Westart with the following one; see, for example, [325, Theorem A].

ℵ Proposition 6.19 Every (metrizable) tvs (E, τ) such that dim E ≥ 2 0 is linearly homeomorphic under a linear map tE to a (metrizable) subspace of a tvs (S(E), μ(τ)) that does not admit nonzero continuous linear functionals and such that dim E equals the codimension of tE(E) (in S(E)) and the density characters of the spaces E and S(E) are the same.

~~We will make use of the following observation due to S. Dierolf [343]; see also [117], [116].~~

~~Lemma 6.9 Let ξ and ϑ be two vector topologies on a vector space E such that ξ ≤ ϑ. If F is a vector subspace of E such that ξ|F = ϑ|F and ξ/F = ϑ/F, then ξ = ϑ.~~

Proof Let U be a ϑ-neighborhood of zero in E. Then there exists a neighborhood of zero V in ξ such that (V − V)∩ F ⊂ U. Since U ∩ V is a neighborhood of zero in ϑ, there exists a ϑ-neighborhood of zero W such that W ⊂ V, W ⊂ (U ∩ V)+ F. Choose arbitrary w ∈ W . Then there exist x ∈ U ∩V and y ∈ F such that w = x +y, where y := −x + w ∈−(U ∩ V)+ W ⊂ (V − V)∩ F ⊂ U. Hence w = x + y ∈ (U ∩ V)+ U ⊂ U + U. This proves that W ⊂ U + U,soU + U is a neighborhood of zero in ξ. Hence ξ = ϑ. 6.8 More examples of nonseparable weakly analytic tvs 177

~~Also, the following lemma will be used; see [343, Proposition 2.1, Theorem 3.2] and [186].~~

~~Lemma 6.10 Let E be a tvs containing a closed vector metrizable (normed) subspace F such that the quotient E/F is also metrizable (normed). Then E is metrizable (normed).~~

Proof Let q : E → E/F be the quotient map. By the assumptions there exists a sequence (Vn)n of neighborhoods of zero in E such that for each n ∈ N we have Vn+1 −Vn+1 ⊂ Vn, (F ∩Vn)n is a basis of neighborhoods of zero in F , and (q(Vn))n is a basis of neighborhoods of zero in the quotient space E/F. Choose neighborhoods of zero U and W in E such that W + W ⊂ U. There exist n, m ∈ N, m>n, such that

~~F ∩ Vn ⊂ W, q(Vm) ⊂ q(W ∩ Vn+1).~~

~~Then Vm ⊂ W ∩ Vn+1 + F . This implies that~~

~~Vm ⊂ W ∩ Vn+1 + F ∩ (Vm − Vn+1) ⊂~~

~~W + F ∩ (Vn+1 − Vn+1) ⊂ W + F ∩ Vn ⊂ W + W ⊂ U. This proves that E has a countable basis of neighborhoods of zero yielding the metrizability of E. The other case we prove similarly.~~

~~Lemma 6.11 can be found in [344, Proposition 12.20].~~

~~Lemma 6.11 Let F ⊂ E be a separable closed vector subspace of a tvs such that E/F is separable. Then E is separable.~~

~~Proof Let {yn : n ∈ N} be a dense subset in E/F. For each n ∈ N,let{xm,n : m ∈ N} be a dense subset of yn + F . Then the set {xm,n : n, m ∈ N} is dense in E. Hence E is separable.~~

~~We also need the following result due to Frolik [176]; see also [344, Proposition 12.21].~~

~~Lemma 6.12 Let E be a metrizable and separable tvs containing a closed vector subspace F such that F and E/F are Baire spaces. Then E is a Baire space.~~

~~Proof Let (Un)n be a countable basis of open sets in E. Assume that E is covered by a sequence (An)n of nowhere dense closed subsets of E.Letq : E → E/F be the quotient map. For each n, m ∈ N,set~~

~~−1 Kn,m := {y ∈ E/F :∅ =Un ∩ q (y) ⊂ Am}. −1 Clearly, each Kn,m has a void interior (since Un ∩ q (int Kn,m) ⊂ Am). 178 6 Weakly Analytic Spaces~~

~~We prove that all sets Kn,m are nowhere dense in E/F. Clearly, Kn,m are closed in q(Un) since~~

~~−1 q(Un) \ Kn,m ={y ∈ E/F : Un ∩ q (y) ∩[E \ Am] =∅}~~

= q (Un ∩ (E \ Am)) . ⊂ Therefore, Kn,m q(Un) is nowhere dense in the open set q(Un). This yields that := Kn,m is nowhere dense in E/F. Set K n,m Kn,m. Since, by the assumption, the quotient space E/F is a Baire space, there exists z ∈ (E/F ) \ K. Clearly, −1 −1 q (z) = An ∩ q (z). n

~~Since q−1(z) is also a Baire space, there exist p,r ∈ N such that~~

~~−1 −1 ∅ =Up ∩ q (z) ⊂ Ar ∩ q (z).~~

~~This implies that z ∈ Kp,r ⊂ K, a contradiction. Hence E is a Baire space.~~

~~We recall the following result; see [65, Chapter III, 3, Exercise 9], or [343, Propo- sition 1.3] or [328, Proposition 2.4.2].~~

~~Lemma 6.13 Let E be a tvs containing a complete vector subspace F such that the quotient space E/F is complete. Then E is complete.~~

Proof Let H be the completion of E. Assume that x ∈ H \ E. Set G := span{E ∪ {x}}.Letqx : H → E/F be the unique extension of the quotient map q : E → E/F. Since qx(x) ∈ E/F, there exists y ∈ E such that qx(x) = q(y). Since F is closed in −1 ∩ = − ∈ ∈ G and qx (0) G F, we deduce that x y F ,sox E, a contradiction. This proves that E is complete.

~~We also need the following [117, (2), p. 194].~~

Lemma 6.14 Let F be a dense linear subspace of a tvs (E, τ), and let ξ be a vector topology on the quotient space E/F. If η denotes the initial topology on E with respect to the canonical injection J : E→ (E, τ) and the canonical surjection Q : E → (E/F, ξ), then η|F = τ|F, η/F = ξ.

Proof Let U be a neighborhood of zero in (E/F, ξ). Then F ⊂ Q−1(U). Therefore η|F = τ|F . Since Q : (E, η) → (E/F, ξ) 6.8 More examples of nonseparable weakly analytic tvs 179 is continuous, ξ ≤ η/F. To prove the equality, ﬁx a neighborhood of zero U in (E/F, η/F ). Then there exists a neighborhood of zero V in (E, τ) and a neighborhood of zero W in (E/F, ξ) such that

~~V ∩ Q−1(W) ⊂ Q−1(U).~~

~~By density, we have V + F = E (i.e., Q(V ) = E/F). Therefore, if z ∈ W , then there exists v ∈ V such that Q(v) = z. Then~~

~~v ∈ V ∩ Q−1(W) ⊂ Q−1(U).~~

~~This implies that z = Q(v) ∈ U. Hence W ⊂ U, and this yields the equality of the topologies.~~

Now we are ready to prove the following general example providing many metrizable and separable nonanalytic dual-separating tvs’s whose weak topologies are analytic. Clearly, such spaces are necessarily nonlocally convex; see [236].

Example 6.5 For every inﬁnite-dimensional separable Fréchet space (E, τ), there exist two metrizable nonanalytic and weakly analytic vector topologies ξ1, ξ2 such that (1) τ = inf{ξ1,ξ2}, (2) ξ1 is Baire and separable and ξ2 is not separable, and (3) (E, τ) = (E, ξ1) = (E, ξ2) (i.e., the three topologies have the same weak topology).

~~Proof Let (xt )t∈T be a Hamel basis of E. Consider a partition (Tn)n of T such that card T = card Tn for all n ∈ N. Set n En := span xt : t ∈ Ti . i=1~~

~~Then the sequence (En)n covers E, and~~

~~ℵ0 dim E = dim En = dim(E/En) = 2 for n ∈ N. By the Baire category theorem, there exists a dense Baire subspace F := p p p Em of E.For0~~

~~τ<ξ1,ξ1/F = α, ξ1|F = τ|F. 180 6 Weakly Analytic Spaces~~

Then ξ1 is metrizable and separable by Lemma 6.10 and Proposition 6.11. The space (E, ξ1) is nonanalytic by Theorem 5.1 applied to the identity map from (E, τ) onto (E, ξ1). Note that (E, ξ1) is a Baire space by Lemma 6.12. Now we construct the topology ξ2. Since (E, ξ1) is a Baire space, the same argument as above applies to choosing a ξ1-dense subspace G of E such that ℵ dim G = dim(E/G) = 2 0 .

ℵ By Proposition 6.19, there exists a 2 0 -dimensional nonseparable metrizable tvs Z without nonzero continuous linear functionals. We proceed as above to deﬁne on E a nonseparable metrizable vector topology ξ2 such that

~~τ<ξ2,τ|G = ξ2|G, and (E/G, ξ2/G) is linearly homeomorphic to the space Z. Clearly, τ ≤ inf{ξ1,ξ2} and~~

~~τ|G = inf{ξ1,ξ2}|G = ξ2|G.~~

On the other hand, the topologies τ/G= ξ1/G are trivial, so the topology τ/G and the topology inf{ξ1,ξ2}/G coincide. By Lemma 6.9, we note τ = inf{ξ1,ξ2}. Finally, we prove that the topologies τ , ξ1 and ξ2 have the same continuous linear functionals on E. This will show that the corresponding weak topologies for the spaces (E, τ), (E, ξ1) and (E, ξ2) are the same. Consequently, each E endowed with the weak topology of (E, ξi), i = 1, 2, respectively, will be analytic. Indeed, let f ∈ (E, ξ1) be a ξ1-continuous linear functional on E, and let h ∈ (E, τ) be an extension of f |F in τ . Since f − h ∈ (E, ξ1) and (f − h)(F ) ={0}, we note that the map x + F → (f − h)(x) belongs to (E/F,ξ1/F ) , so h(x) = f(x) for each x ∈ E (i.e., f = h ∈ (E, τ) ). The same proof works for the topology ξ2.

~~Since every metrizable K-analytic space is analytic, Example 6.5 can be used to deduce the following example.~~

Example 6.6 Every inﬁnite-dimensional separable Fréchet space E admits a strictly ﬁner metrizable and separable vector topology ξ with the same continuous linear functionals as the original one of E and such that (E, ξ) is a Baire space that is not K-analytic.

~~Saxon [355] asked if Theorem 2.4 remains true for nonlocally convex spaces. We note the following observation by using the argument from Example 6.5.~~

Corollary 6.17 Assume CH. If (E, ξ) is a tvs containing a dense inﬁnite-codimensional subspace F , then E admits a stronger vector topology υ such that (E, ξ) = (E, υ) and (E, υ) contains a dense non-Baire hyperplane. 6.8 More examples of nonseparable weakly analytic tvs 181

ℵ Proof By the assumption, the codimension of F in E is either 2 0 or ℵ0.Fix0< p<1. Then the quotient space E/F admits a ﬁner separable vector topology μ such that in the ﬁrst case (E/F, μ) is isomorphic to the space Lp and in the other p case (E/F, μ) is isomorphic to a dense ℵ0-dimensional vector subspace of L .As in Example 6.5, there exists on E a stronger vector topology υ such that ξ|F = υ|F and υ/F = μ. By Remark 2.1, it follows that (E/F, μ) contains a dense non-Baire hyperplane. Then, as we showed in Theorem 2.4, the space E contains a dense non- Baire hyperplane. Chapter 7 K-analytic Baire Spaces

Abstract In this chapter, we show that a tvs that is a Baire space and admits a countably compact resolution is metrizable, separable and complete. We prove that N a linear map T : E → F from an F-space E having a resolution {Kα : α ∈ N } into atvsF is continuous if each restriction T |Kα is continuous. This theorem (due to Drewnowski) was motivated by the Arias–De Reina–Valdivia–Saxon theorem about non-Baire dense hyperplanes in Banach spaces. We provide a large class of weakly analytic metrizable and separable Baire tvs that are not analytic (clearly such spaces are necessarily not locally convex).

~~7.1 Baire tvs with a bounded resolution~~

We know already from Corollary 3.12 that a Baire lcs that is a quasi-(LB)-space is a Fréchet space. Tkachuk [399] proved that if Cp(X) is K-analytic and Baire, the space X is countable and discrete. Hence a K-analytic Baire space Cp(X) is a separable Fréchet space. In fact, Tkachuk’s theorem follows from Theorem 7.1 below due to De Wilde and Sunyach [111]; see also Valdivia [421, p. 64].

~~Theorem 7.1 (De Wilde–Sunyach) A Baire K-analytic lcs is a separable Fréchet space.~~

Lutzer, van Mill and Pol [279] exhibited a countable space X (having a unique nonisolated point) such that Cp(X) is a separable, metrizable, noncomplete Baire space that is not K-analytic. In this section, we prove Theorem 7.2 from [233], which extends Theorem 7.1.

~~Theorem 7.2 (Kakol–López-Pellicer) ˛ ABairetvsF with a relatively countably compact resolution is a separable metrizable and complete tvs.~~

~~For the proof, we need two additional results from [233].~~

~~Proposition 7.1 Every Baire tvs with a bounded resolution is metrizable. Any metrizable tvs has a bounded resolution.~~

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 183 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_7, © Springer Science+Business Media, LLC 2011 184 7 K-analytic Baire Spaces

N Proof Let {Kα : α ∈ N } be a bounded resolution in E. As usual, for α = ∈ NN := { : = = = } := (nk) ,setCn1,n2,...,nk Kβ β (ml), nj mj ,j 1,...,k and Wk ∈ N ∈ N Cn1,n2,...,nk , k . Then, for every neighborhood of zero U in E, there exists k such that k Wk ⊂ 2 U. Indeed, otherwise there exists a neighborhood of zero U in E such that for every −k k ∈ N there exists xk ∈ Wk such that 2 xk ∈/ U. Since xk ∈ Wk for every k ∈ N, = k ∈ NN there exists βk (mn)n such that ∈ = k xk Kβk ,nj mj , = = { k : ∈ N} ∈ N = for j 1, 2,...,k. Set an max mn k for n , and set γ (an). Since ≥ ∈ N ⊂ ∈ ∈ N γ βk for every k , then Kβk Kγ . Hence xk Kγ for all k .ThesetKγ −k is bounded, so 2 xk → 0inE, which provides a contradiction. Since E is a Baire space and = = E Cn1 ,Cn1 Cn1,n2 ,..., n1 n2 N there exist sequences (nk) ∈ N , (xk)k in E, and a sequence (Uk)k of neighborhoods of zero in E such that

xk ∈ int Wk,xk + Uk ⊂ Wk, for all k ∈ N. Choose arbitrary closed, balanced neighborhoods of zero U and V in k E such that V + V ⊂ U. Since there exists k ∈ N such that Wk ⊂ 2 V , we note that −k −k −k 2 Uk ⊂ 2 Wk − 2 xk ⊂ V + V ⊂ U. −k This proves that the sequence (2 Uk)k forms a countable basis of neighborhoods of zero in E,soE is metrizable. Now assume that E is a metrizable tvs. Let (Un)n be a countable basis of bal- N anced neighborhoods of zero for E.Forα = (nk) ∈ N ,setKα := nkUk. It is N k easy to see that {Kα : α ∈ N } is a bounded resolution in E.

It is known, that if F is a Cech-completeˇ space and E is a completely regular Hausdorff space containing F as a dense subspace, E \ F is of ﬁrst Baire category; see [343, Corollary 13.5]. Making use of Theorem 6.1, we note that if F is a metric space having a compact resolution swallowing compact sets, F is Cech-complete.ˇ We provide another applicable result of this type for Baire spaces F admitting a certain resolution.

Proposition 7.2 Let E be a topological space that admits a weaker topology ξ generated by a metric d. Let F be a dense Baire subset of E having a resolution N {Kα : α ∈ N } consisting of closed sets in ξ. Then E \ F is of ﬁrst Baire category.

Proof We claim that ⊂ O(Cn1,n2,...,nk ) F (7.1) k 7.1 Baire tvs with a bounded resolution 185 = ∈ NN ∈ ∈ ∈ N for α (nk) . Indeed, if z k O(Cn1,n2,...,nk ), z E, for every k there exists ∈ −1 ∩ xk B(z,k ) Cn1,n2,...,nk , where B(z,k−1) denotes an open ball in (E, d) with its center at point z and ra- −1 dius k . This implies that the sequence (xk)k converges to z in the topology ξ.On N the other hand, there exists γ ∈ N such that xk ∈ Kγ for all k ∈ N. Since Kγ is closed in (F, d),wehavez ∈ F . This proves the claim. Now define the sets := { : ∈ N} K0 O(F) O(Cn1 ) n1 , := { : ∈ N} ≥ Kn1,...,nk−1 O(Cn1,...,nk−1 ) O(Cn1,...,nk−1,nk ) nk ,k 2, and recall that E = O(F). Applying Proposition 2.5, we note that every set of the type above is nowhere dense. Note also that \ ⊂ ∪ E F K0 Kn1,...,nk . (7.2) Indeed, to prove this fact, assume that there exists ∈ ∪ x/K0 Kn1,...,nk . ∈ N ∈ ∈ N Therefore there exists n1 such that x O(Cn1 ). Next, we find n2 such ∈ = that x O(Cn1,n2 ). Following this inductive procedure, we obtain a sequence α ∈ NN ∈ ∈ N (nk) such that x O(Cn1,n2,...,nk ) for every k .By(7.1), we derive that x ∈ F . This proves the inclusion (7.2). From (7.2), it follows that indeed the space E \ F is of first Baire category in E.

N If {Kα : α ∈ N } is a compact resolution, the argument is easier: By Corol- lary 3.6, the space F is K-analytic. From [421, Chapter 1, 4.3(19)], we note that F has the Baire property in E. Note also that E = O(F) since all nonvoid open sets in E intersect F in a set of second Baire category. Now, from the Baire property of F in E, it follows that E \ F(= O(F)\ F) is of ﬁrst Baire category in E;see[261, Corollary 2,11.IV] or Proposition 2.4. Now we are ready to prove Theorem 7.2.

Proof Since in a tvs relatively countably compact sets are bounded, we apply Propo- sition 7.1 to deduce that F is metrizable. Consequently, a metrizable F has a compact resolution, and by Corollary 6.2 the space F is analytic and hence separable. Let E be the completion of F . By Proposition 7.2,thesetE \ F is of ﬁrst Baire category. We prove that E = F .IfE = F , then taking x ∈ E \ F we would have x + F ⊂ E \ F, and this provides a contradiction since x + F and E \ F are of second and ﬁrst Baire category, respectively. Hence E is separable, metrizable and complete. 186 7 K-analytic Baire Spaces

~~Applying Proposition 7.2 and using a similar argument as in the proof of Theo- rem 7.2, we note the following corollary.~~

~~Corollary 7.1 Let E be a metrizable Baire tvs having a complete resolution. Then E is complete.~~

~~Corollary 7.2 below supplements Theorem 7.2 and Canela’s corresponding result from [79].~~

Corollary 7.2 Let X be a paracompact and locally compact space. The following assertions are equivalent: (i) Cp(X) has a bounded resolution. (ii) X is σ -compact. (iii) Cc(X) is a Fréchet space. Moreover, if X is metric and locally compact, Cp(X) has a bounded resolution if and only if Cp(X) is analytic.

Proof Since X is paracompact and locally compact, X is a topological direct sum of a disjoint family {Xi : i ∈ I} of locally compact σ -compact spaces. Hence Cp(X) = i∈I Cp(Xi). (i) ⇒ (ii): If Cp(X) has a bounded resolution, the set I is countable. Indeed, RA otherwise i∈I Cp(Xi) would contain a closed subspace of the type for some uncountable A. Since RA is a Baire space, we apply Theorem 7.2 to deduce that RA is metrizable. Consequently, the set A is countable. This shows that X is σ -compact. The implications (ii) ⇒ (iii) ⇒ (i) are obvious. The last part of the corollary follows from the previous one and a known fact stating that a continuous image Cp(X) of a separable Fréchet space Cc(X) is analytic.

Theorem 7.2 fails in general for topological groups. Any nonmetrizable compact topological group provides such an example. Nevertheless, Proposition 7.2 is applied to prove the following variant of Theorem 7.2;see[99, Theorem 5.4].

~~Theorem 7.3 (Christensen) A Baire topological group E that is an analytic space is a Polish space.~~

Proof Let T be a continuous surjection of NN onto E.LetT be the set of all x ∈ NN for which there exists a neighborhood U(x) of x such that T(U(x))is of first Baire category. Note that T = NN. Indeed, otherwise (since NN is Lindelöf) there exists N a sequence (U(xn))n of neighborhoods of the points xn covering N and such that T(U(xn)) is of first Baire category. Also, n T(U(xn)) is of first Baire category. On the other hand, N T(N ) = E = T(U(xn)) n is the space of second Baire category, a contradiction. This proves that there exists y ∈ NN such that for each neighborhood U(y) of y the set T(U(y)) is of second 7.2 Continuous maps on spaces with resolutions 187

N Baire category. Let (Vn(y))n be a basis of closed neighborhoods of y in N .Note that −1 {T(Vn(y))T (Vn(y)) : n ∈ N} is a basis of neighborhoods of the unit element e of E. Since T(Vn(y)) is of second Baire category and has the Baire property for each n ∈ N (note that T(Vn(y)) is −1 analytic and therefore has the Baire property), each set T(Vn(y))T (Vn(y)) is a neighborhood of e; compare the proof of Proposition 2.7. Take neighborhoods U and V of e in E such that VV−1 ⊂ U. By the continuity of T ,wehave

T(Vk(y)) ⊂ VT(y) for some k ∈ N. Hence −1 −1 −1 T(Vk(y))T (Vk(y)) ⊂ V T (y)T (y) V ⊂ U. This condition for topological groups yields the metrizability. Since E is Lindelöf and hence separable, there exists a Polish group F that contains E as a dense subset. By Proposition 7.2,thesetF \ E is of ﬁrst Baire category. Assume that there exists z ∈ F \ E. Then zE ⊂ F \ E is also of ﬁrst Baire category. As E is of second Baire category, the same holds for zE, a contradiction. Hence E = F is a Polish space.

We complete this section by showing that Proposition 7.1 for locally convex spaces can also be deduced from Corollary 3.12. Indeed, let E be a Baire lcs hav- N ing a bounded resolution {Aα : α ∈ N }. We may assume that all bounded sets Aα N are absolutely convex. For every α = (nk) ∈ N ,letKα be the closed, absolutely convex hull of Aα in the completion F of E.LetG be the linear span (in F )of N {Kα : α ∈ N }. Fix arbitrary x ∈ G. Then there exist scalars tp ∈ R for 1 ≤ p ≤ n, ∈ NN ∈ ≤ ≤ αp and xαp Kαp for 1 p n such that n = ∈| | x tpxαp t nKγ , p=1 where

|t|=max{|tp|:p = 1, 2,...n} N and γ ∈ N such that αp ≤ γ for every p = 1, 2,...n. Choose m1 ∈ N, β = (mk) ∈ N N , such that n|t|≤m1 and γ ≤ β. Then x ∈ m1Kβ . Therefore N G = {n1Kα : α = (nk) ∈ N }. Hence G is a Baire locally convex space having a resolution of Banach discs. By Corollary 3.12, the space G is metrizable and complete. Hence E is metrizable.

~~7.2 Continuous maps on spaces with resolutions~~

~~A possible approach to prove a deep theorem like Theorem 2.4 for separable inﬁnite-dimensional Fréchet spaces might be related to the following problem posed in [233]. 188 7 K-analytic Baire Spaces~~

Problem 7.1 Let E be an inﬁnite-dimensional separable Fréchet space. Does there N exist on E a compact resolution {Kα : α ∈ N } and a discontinuous linear functional N λ on E that is λ|Kα continuous for each α ∈ N ?

We know already that a separable Fréchet space E always admits a compact resolution (even swallowing compact sets). Assume for a moment that Problem 7.1 has a positive answer. Let H := ker λ be the (dense) hyperplane of E deﬁned by λ.Note N that by the assumption the family {Kα ∩ H : α ∈ N } is a compact resolution in H . Then, by Theorem 7.2, the hyperplane H cannot be a Baire space. This approach motivates Problem 7.1. It turns out that the answer to Problem 7.1 has a negative solution, recently proved by Drewnowski [131]. In this section, we prove a couple of results (mostly due to Drewnowski [131]). Theorem 7.4 extends the earlier Corollary 5.3, proved by using Theorem 5.1. Recall that an F-space is a metrizable and complete tvs.

Theorem 7.4 (Drewnowski) Let E be an F-space having a resolution {Aα : α ∈ NN}. Let f : E → F be a linear map from E into a tvs F such that each restriction f |Aα is continuous. Then f is continuous on E.

~~For the proof, we need the following three additional lemmas from [131].~~

N Lemma 7.1 Let E be a topological space having a resolution {Aα : α ∈ N }. Let f : E → F be a map from E into a regular topological space F . If each f |Aα is continuous, f is continuous on the closure Kα of each Aα.

N Proof Fix Aα and x ∈ Kα. There exists β ∈ N such that Aα ⊂ Aβ and x ∈ Aβ . Since f |Aβ is continuous at x,themapf |(Aα ∪{x}) is also continuous at x.This is just enough to claim that f |Kα is continuous; see [146, Exercise 3.2(A)(b)].

~~Lemma 7.2 is easy to check; the proof is left to the reader.~~

N Lemma 7.2 Let E and F be topological spaces and {Aα : α ∈ N } a resolution on E. Assume that f : E → F is a map continuous on each Aα. Then: N N (i) If the resolution {Aα : α ∈ N } is compact, {f(Aα) : α ∈ N } is a compact resolution on the range space f(E). N (ii) If the resolution {Aα : α ∈ N } is compact (closed), for every closed set −1 N C ⊂ F the family {f (C) ∩ Aα : α ∈ N } is a compact (closed) resolution on the subset f −1(C) of E.

Lemma 7.3 Assume that a linear map f : E → F is continuous on a closed set X ⊂ E and on a compact set K ⊂ E. Let A be a compact set of nonzero scalars. Then f is continuous on the closed set AX + K.

Proof Let (zt )t∈T be a net in AX + K converging to z ∈ E. For each t ∈ T , there exist xt ∈ X, at ∈ A and yt ∈ K such that zt = at xt + yt . By the compactness of A 7.2 Continuous maps on spaces with resolutions 189

~~−1 and K, we may assume that at → a ∈ A, yt → y ∈ K. Then xt → a (z − y) := x ∈ X. Hence z ∈ AX + K. Finally, we note that~~

~~f(zt ) = at f(xt ) + f(yt ) → af (x ) + f(y)= f(z). Hence f is continuous at the point z.~~

~~Now we are ready to prove Theorem 7.4.~~

Proof Applying Lemma 7.1, we may assume that each set Aα is closed in E.Ob- serve that the kernel N := {x ∈ E : f(x)= 0} is a closed subspace of E. Indeed, without loss of generality, we may assume that N is dense in E. We check that N = E by showing that f(E)⊂ V0 for each closed, balanced neighborhood of zero V0 in F . Let (Vn)n be a sequence of closed, balanced neighborhoods of zero in F such that

Vn+1 + Vn+1 ⊂ int Vn ∈ N ∪{ } := −1 = for each n 0 . Set Un f (Vn). Clearly, U n≥0 Un is a linear subspace of E. Since N ⊂ U, U is dense, and from Lemma 7.2 (ii) it follows that −1 each Un admits a closed resolution. Clearly, f (int Vn) is a dense, algebraically open subset of E. Then, for every nonempty open subset G of E, the intersection −1 −1 G ∩ f (int Vn) is of second Baire category in E. Therefore f (int Vn) and Un are Baire and dense subspaces of E. Applying Proposition 7.2, we note that each E \ Un is of first Baire category. This shows that E \ U = E \ Un = E \ Un n n is of first Baire category. If y ∈ E \ U, then y + U ⊂ E \ U, and this shows that the Fréchet space E = (E \ U)∪ U is of first Baire category, a contradiction. Hence f(E)⊂ V0. This proves that the kernel of f is closed. Next, we prove that the map f is continuous. Assume first that F is an F-space, and define the linear map T : E × F → F, T (x,y) := y − f(x). Consider the resolution N {Aα × F : α ∈ N } on the F-space E × F . Since ker (T ) = Graph (f ) is a closed subspace in E × F by the last observation, we use the closed graph theorem between F-spaces [1, Theo- rem 8.6] to deduce that T is continuous. Hence fis continuous as well. Finally, every tvs is a subspace of a product i∈I Fi of F-spaces (see [1, The- orem 3.5]). Using the previous case for f composed with each of the canonical → ∈ projections i∈I Fi Fi , which is clearly continuous for each i I , we deduce that f is continuous. This completes the proof.

~~Theorem 7.4 is applied to prove the following corollaries; see [131]. 190 7 K-analytic Baire Spaces~~

Corollary 7.3 Let E be an F-space, and let E0 be a subset of E admitting a closed resolution. Let (Ej )1≤j≤n be a ﬁnite family of subsets of E admitting a compact = + : → resolution. If E E0 1≤j≤n Ej and a linear map f E F to a tvs F is continuous on each set Ej for 0 ≤ j ≤ n, then f is continuous on E.

j N Proof For each 0 ≤ j ≤ n,let{Kα : α ∈ N } be a resolution on the set Ej . Set := 0 + 1 +···+ n Kα Kα Kα Kα . N Then {Kα : α ∈ N } is a closed resolution on E, and by Lemma 7.3 the map f is continuous on each set Kα. Now Theorem 7.4 applies for the continuity of f .

Corollary 7.4 Let E be an F-space equal to an algebraic direct sum of vector subspaces (Ej )0≤j≤n. Assume that E0 admits a closed resolution while the remaining subspaces admit a compact resolution. Then the associated projections pj : E → Ej are continuous. Hence all subspaces Ej are closed in E.

~~Corollary 7.5 If a linear subspace F of an F-space E has countable codimension in E and admits a closed resolution, F is closed and has ﬁnite codimension.~~

Proof Let G be a countable-dimensional vector subspace of E such that E = F ⊕G (algebraically). Clearly, G admits a compact resolution as an analytic space. We apply Corollary 7.4 to deduce that both spaces F and G areclosedinE. By Corol- lary 2.3, we conclude that G is ﬁnite-dimensional.

~~The proofs of the next two simple corollaries are left to the reader.~~

Corollary 7.6 Let E be a separable F-space having a compact resolution {Kα : α ∈ NN}. Then, for every (sequentially) complete tvs F , the space L(E, F ) endowed with the topology of uniform convergence on sets Kα is (sequentially) complete.

N Corollary 7.7 Let E be an F-space having a resolution {Aα : α ∈ N }. Then the original topology of E coincides with the ﬁnest vector topology ξ for which all inclusions jα : Aα → (E, ξ) are continuous.

Corollary 7.7 shows that for a separable F-space E there does not exist a strictly ﬁner vector topology ξ such that (E, ξ) has a compact resolution. This observation follows also from Theorem 5.1. A simple application of Theorem 7.4 says that a linear map f : E → F from an F-space E to tvs F that is continuous on an absorbing subset A ⊂ E is continuous on the whole space E. Indeed, set An := nA, n ∈ N, := = ∈ NN and then Aα An1 for α (nk) . We complete this section with the following observation; see [131].

Proposition 7.3 Let E be a metrizable lcs covered by an increasing sequence (Kn)n of compact sets. Assume that every linear functional on E that is continuous on each set Kn is continuous. Then E is ﬁnite-dimensional. 7.2 Continuous maps on spaces with resolutions 191

Proof By EK we denote the set of all linear functionals f on E that are con- = tinuous on each set Kn. Since E EK , the space E endowed with the topology ξK of the uniform convergence on the sets (Kn)n is a Fréchet space. Clearly,

σ(E ,E)≤ ξK . The space E is metrizable, and hence there exists a countable basis ◦ (Un)n of closed, absolutely convex neighborhoods of zero of E. Since each polar Un = ◦ of Un is σ(E ,E)-compact, it is closed in ξK . Since E n Un , the Baire category ∈ N ◦ ⊂ ◦ ∈ N theorem applies to get m such that tKm Un for some t>0 and m .This −1 implies that Un ⊂ t acKm, which implies that E has a precompact neighborhood of zero. Hence E is ﬁnite-dimensional; see, for example, [213, Theorem 3.5.6]. Chapter 8 A Three-Space Property for Analytic Spaces

Abstract In this chapter, we show that analyticity is not a three-space property. We prove, however, that a metrizable tvs E is analytic if it contains a complete locally convex analytic subspace F such that the quotient E/F is analytic. We reprove (using Corson’s example) that the Lindelöf property is not a three-space property.

~~8.1 An example of Corson~~

By a three-space property (for topological vector spaces) we understand the following [343]: Suppose E is a tvs and F ⊂ E is a closed vector subspace of E such that F and the quotient E/F have a certain property P. Does E have property P? Several important topological properties of tvs’s are three-space properties such as separability, local boundedness, completeness, or barelledness; see [344], [343] and Lemmas 6.10, 6.11, 6.12, 6.13. In this section, we provide an interesting example, originally due to Corson [103, Example 2], showing that the Lindelöf property is not a three-space property. We shall need several additional partial results of independent interest; see also [149, Theorem 12.43]. Let us brieﬂy recall the main idea related to Corson’s example. Let E be the subspace of the space ∞[0, 1] formed by all bounded real-valued functions on [0, 1] that are right-continuous and have ﬁnite left limits. The space (E, σ (E, E)) is not normal. Hence E is not weakly Lindelöf. The space E contains a subspace isomorphic with C[0, 1], and the quotient E/C[0, 1] is isomorphic to the WCG Banach space c0[0, 1], so the weak topology of c0[0, 1] is K-analytic. Now we start with some facts in order to present Corson’s example. We need the following simple fact; see [146, Problem 2.7.12(b)].

~~Lemma 8.1 Let I beaset. Let U and V be disjoint open subsets in KI . Then there exists a countable set J ⊂ I such that pJ (U) ∩ pJ (V ) =∅, where pJ is the projection of KI onto KJ .~~

~~We also need the following lemma due to Corson and Isbell [102, Theorem 2.1].~~

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 193 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_8, © Springer Science+Business Media, LLC 2011 194 8 A Three-Space Property for Analytic Spaces

~~Lemma 8.2 Let I be a set. Let W ⊂ KI be a dense subset. For every continuous function f : W → K, there exist a countable J ⊂ I and a continuous function g : pJ (W) → K such that f = g ◦ pJ .~~

Proof For each rational number r,set −1 −1 Ur := f (r, ∞), Vr := f (−∞,r). 1 1 KI = 1 ∩ = Choose open disjoint sets Ur and Vr in such that Ur Ur W and Vr 1 ∩ ⊂ Vr W . Lemma 8.1 provides countable Jr I such that 1 ∩ 1 =∅ pJr (Ur ) pJr (Vr ) . := = ∈ = ∈ = Set I0 r Jr . Note that if x (xi)i∈I W , y (yi)i∈I W , and xi yi for i ∈ I0, then f(x)= f(y).

~~This is applied to get the following interesting lemma [103, Example 2].~~

~~Lemma 8.3 Let E be an lcs. Let F be the set of all continuous real-valued functions ℵ on (E, σ (E, E )). Then |F|≤2 0 dim (E ).~~

Proof Clearly, (E∗,σ(E∗,E)) is linearly homeomorphic with KI , where I is a set with |I|=dim E and E∗ denotes the algebraic dual of E. Note that E is identiﬁed ∗ ℵ with a dense subspace of E . By Lemma 8.2,wehave|F|≤2 0 dim(E ) m, where ℵ m = 2 0 denotes the cardinality of continuous real-valued functions on separable ℵ R 0 .

ℵ By c0(I) we denote the Banach space (for |I|=2 0 ) of functions f : I → K such that for each ε>0 there exists a ﬁnite set J ⊂ I such that |f(i)| <εfor all i ∈ I \J . Note that c0(I) is a WCG Banach space (i.e., c0(I) contains a weakly compact set whose linear span is dense in c0(I)). Indeed, let A := (et )t∈I be the family of the standard unit vectors in c0. Observe that the set {et ∪{0}:t ∈ I} is weakly compact in c0. Indeed, each sequence (en)n in A of distinct elements converges pointwise to zero, and hence (en)n converges to zero weakly in c0. This means, by applying the classical Eberlein–Šmulian theorem, that A is weakly compact in c0. Since c0(I) is a WCG Banach space, we refer the reader to Theorem 12.8 below (and Proposition 3.4; see also [103, Theorem 1]).

~~Lemma 8.4 The space c0(I) is weakly K-analytic and hence weakly Lindelöf.~~

~~Now we are ready to prove the following theorem due to Corson [103, Exam- ple2];seealso[124].~~

Theorem 8.1 (Corson) There exists a Banach space E such that E is not normal in the weak topology and contains a closed subspace F such that F and E/F are weakly Lindelöf. 8.1 An example of Corson 195

Proof Since every regular Lindelöf space is normal by Lemma 6.1, the space E is not weakly Lindelöf. Let E be the subspace of ∞[0, 1] formed by all bounded real- valued functions on [0, 1] that are right-continuous and have ﬁnite left limits. Endow := | | ∈ E with the norm x supt∈[0,1] x(t) for all x E.LetF be the subspace of E of continuous functions; therefore F is isometric to the space C[0, 1]. Clearly, F in the weak topology is analytic and hence Lindelöf. We show that the quotient space E/F is linearly homeomorphic to the space c0[0, 1]. Set I := [0, 1], and deﬁne continuous linear functionals on E by the formulas

gt (x) := x(t), ht (x) := lim u→t− x(u), for each t ∈ I . Set ft := gt − ht . We show that the map T deﬁned by T(x):= (ft (x))t∈I is a continuous linear surjection from E onto c0(I). Note that T is linear, T(E)⊂ c0(I) and ker T = F . Clearly,

|ft (x)|=|x(t) − lim u→t− x(u)|≤2x. This yields T(x)≤2x,soT is continuous. Now we show that T is a surjection. Fix arbitrary (yt )t∈I in c0(I). We construct a sequence (xn)n∈N∪{0,−1} in E that uniformly converges. Deﬁne ﬁnite sets for n ≥ 1 −n 1−n I0 := {t ∈ I : 1 ≤|yt |},In := {t ∈ I : 2 ≤|yt | < 2 }.

Deﬁne x−1(t) = 0 for each t ∈ I . Next, we construct the sequence (xn)n as follows: xn(t) = xn−1(t), lim xn(u) = lim xn−1(u), t ∈ Ik, u→t− u→t− kn m−1 −k −n+1 |xm(t) − xn(t)|≤ 2 < 2 . k=n

This shows that (xn)n is a Cauchy sequence in E, so its limit x ∈ E satisﬁes T(x)= (yt )t∈I . By the closed graph theorem, we deduce that E/F is linearly homeomorphic to the space c0(I). Now we prove that E is not normal in its weak topology. Set 0 tt + := = (gt ht )(xu) ⎩1 u t 0 u

Note also that A is closed. On the other hand, it is known that on a closed dis- ℵ crete set of cardinality c := 2 0 there exist 2c continuous real-valued functions. As- sume that (E, σ (E, E)) is normal. Every such function from A can be extended by the Tietze–Urysohn theorem to the space (E, σ (E, E)). To get a contradiction, ℵ it is enough to show that there exist only 2 0 continuous real-valued functions on ℵ the space. This is the case if we realize that dim E ≤ 2 0 by using Lemma 8.3. ℵ The last inequality follows from the well-known inequalities dim F ≤ 2 0 and ℵ dim(E/F ) ≤ 2 0 .

~~We shall come back to Corson’s example in the context of the separable complementation property (see Theorem 18.2 below).~~

~~8.2 A positive result and a counterexample~~

Corson’s example also shows that the K-analyticity is not a three-space property since E is not weakly Lindelöf and the spaces C[0, 1] and E/C[0, 1] (as isomorphic to a WCG Banach space) are weakly K-analytic. On the other hand, the space c0[0, 1] is not separable, so this example does not cover the case for P analytic. If E is a tvs containing a vector subspace F such that F and E/F are separable Fréchet spaces, it is a separable Fréchet space. This is clear since the separability, metrizability and completeness are a three-space property; see Lemma 6.10, Lemma 6.11 and Lemma 6.13. Let us consider two other cases: Let F and E/F be analytic, and let E be metrizable. Assume that F (or E/F) is complete. Is E analytic? In order to provide a positive result concerning a three-space property for the analyticity, we need additional facts; see [264, Theorem 1], [273] and [274].

Proposition 8.1 (Labuda–Lipecki) Assume that E := (E, .) is a separable inﬁ- nite-dimensional Banach space. Then there exists a linearly independent sequence { : ∈ N} (yn)n in E such that n yn converges, span yn n is dense in E and if ∈ ∞ = = ∈ N (tn) and n tnyn 0, then tn 0 for all n .

Proof First observe that if X is a proper subspace of E, the space E \ X is dense in E. Indeed, ﬁx ε>0, and take x ∈ E \ X with x <ε. Then (x + y) − y <ε and x + y ∈ E \ X provided y ∈ X. This will be used to see that E contains a dense linearly independent sequence (xn)n. Indeed, let {zn : n ∈ N} be a dense subset of E. By the remark above, (E \ X) ∩{zn : n ∈ N} is dense in E if X is any ﬁnite-dimensional subspace of E. Then, we can construct by a simple induction a sequence (znk )k with ∈ { } znk+1 / span zn1 ,zn2 ,...,znk − ≤ −1 ∈ N { : ∈ N} and such that znk zk k for all k .Theset znk k is dense in := ∈ N E (by the last inequality) and linearly independent. Set xk znk for all k . 8.2 A positive result and a counterexample 197

−n Take numbers βn > 0 such that βxn≤2 for |β|≤βn and all n ∈ N. Then, the series n βnxn is bounded multiplier convergent (i.e., for each bounded sequence ∈ ∞ ∈ N := (tn)n , the series n tnβnxn converges in E). For each n ,setyn βnxn. We show that (yn)n contains a subsequence (yn )k as desired. Put k n −1 Kn := siyi :|si|≤1,i≤ n − 1, 2 ≤|sn|≤1 , i=0 where n ∈ N and y0 := 0. Note that the sets Kn are compact and 0 ∈/ Kn. Since E \ Kn is an open neighborhood of zero for each n ∈ N, there exists m>nsuch that ∞ tiyi :|ti|≤1,i≥ m ⊂ E \ Kn. i=m

~~Hence, we can choose a sequence n1~~

~~Now we are ready to prove the following theorem [236, Theorem 12].~~

Theorem 8.2 (i) Let E be a metrizable tvs containing a closed subspace F such that F and E/F are analytic. If F is complete and locally convex, then E is analytic. (ii) There is a separable normed space E that is not analytic and contains a closed analytic subspace F such that E/F is a separable Banach space.

Proof Proof of (i) is due to Drewnowski (private communication). Let G be the completion of E.LetQ : G → G/F be the quotient map. By a result of Michael [289](seealso[50, Proposition 7.1 or Corollary 7.1]) for Fréchet spaces, there is a continuous map g : G/F → G such that Q ◦ g is the identity map on G/F , g(x + F)∈ x + F, for each x ∈ G. Since the quotient E/F ⊂ G/F is analytic, the quotient space E/F N N admits a compact resolution {Kα : α ∈ N }. Next, assume that {Aα : α ∈ N } is a compact resolution on F . We show that the compact sets

Mα := g(Kα) + Aα N form a compact resolution on E for α ∈ N . Indeed, first observe that g(Kα) ⊂ E, so then each compact set Mα is contained in E.Fixx ∈ E. Since g(x + F)∈ x + F, 198 8 A Three-Space Property for Analytic Spaces there exists y ∈ F such that g(x + F)+ y = x. Then x + F ⊂ Kα and y ∈ Aα for N N some α ∈ N . This shows that x ∈ Mα. Consequently, E = {Mα : α ∈ N }.We proved that the space E (metrizable and separable) has a compact resolution. Now we apply Corollary 6.2 to deduce that E is analytic. Now we prove (ii). Fix an infinite-dimensional separable Banach space E := (E, .). By Proposition 8.1, we obtain a sequence (yn)n in E such that ∞ { : ∈ N} ∈ ∞ = yn < , span yn n is dense in E and if (tn) with n tnyn 0, then (t ) = 0. Define a compact injective map n 1 T : → E, T (x) := xnyn, n 1 1 where x = (xn) ∈ . Clearly, the image F := T( ) is a dense subspace of E, and since the map T is compact, F is a proper subspace of E. On the other hand, F is not barrelled. Indeed, assume that F is a barrelled subspace of E. By Proposition 2.11 and Proposition 2.12, F is Baire-like. Since T is an injective continuous map, F admits a strictly finer topology γ such that (F, γ ) is linearly homeomorphic with the Banach space 1. The identity map from Baire-like F onto (F, γ ) has a closed graph, so by Theorem 2.7 this map is continuous. Hence γ equals the original topology of F , a contradiction. ℵ ℵ Note that dim(E/F ) = 2 0 . Indeed, since dim E = 2 o , we apply Theorem 2.6 to get the claim. Let τ be a normed topology defined by the norm ., and let q : E → E/F be the quotient map. Since the quotient topology of E/F is trivial and dim 1 = dim E/F, the space E/F admits a stronger separable Banach topology α such that (E/F, α) is isomorphic to 1. Therefore the assumptions of Lemma 6.14 are satisfied. Hence there exists on E a coarsest vector topology ξ such that τ<ξ, ξ/F = α and ξ|F = τ|F . Note that the sets U ∩ q−1(V ), where U and V run over τ - and α-neighborhoods of zero, respectively, form a basis of neighborhoods of zero for ξ. Since the separability and the property of being a normed space are a three-space property (see Proposition 6.10 and Proposition 6.11), the topology ξ is normed and separable. Finally, since by Theorem 5.1 every linear map with a closed graph from a Ba- nach space into an analytic space is continuous, we deduce that the space (E, ξ) is not analytic. Its closed subspace F is analytic, and the quotient space (E/F, α) is isomorphic to 1. Hence α is the topology of a separable Banach space.

~~Applying the argument used in the proof of Theorem 8.2, we note the following result.~~

ℵ Proposition 8.2 Let Y be a WCG Banach space with dim Y = 2 0 ; for example, let Y := c0[0, 1]. Then there exists an lcs E such that E is not a Lindelöf Σ-space, E contains a closed subspace F that is a continuous image of 1 and E/F is linearly homeomorphic to the K-analytic space Y endowed with the weak topology. Consequently, the Lindelöf Σ property is not a three-space property.

Proof Let (E, ξ) be an inﬁnite-dimensional separable Banach space, and let T : 1 → E be an injective compact map such that T(1) is a proper dense subspace 8.2 A positive result and a counterexample 199 of E (see the proof of Theorem 8.2). Note that a similar argument that was used ℵ above yields that dim (E/F ) = 2 0 , where F = T(1). Then the quotient space E/F, whose topology ξ/F is trivial, admits a locally convex nonseparable K- analytic topology γ such that ξ ≤ γ , and (E/F, γ ) is linearly homeomorphic to (Y, σ (Y, Y )). By Lemma 8.4, the space (Y, σ (Y, Y )) is K-analytic. Applying the procedure as in the proof of Theorem 8.2, we obtain a locally convex topology θ on E such that ξ ≤ θ, ξ|F = θ|F, θ/F = γ. Since (E/F, γ ) is nonseparable, (E, θ) is nonseparable. Finally, note that E = (E, θ) is not a Lindelöf Σ-space. Indeed, assume that (E, θ) is a Lindelöf Σ-space. Since ξ is a separable metrizable topology and ξ ≤ θ, by Lemma 3.1 we conclude that (E, θ) is separable, a contradiction.

~~We complete this section with the following.~~

~~Problem 8.1 Is a metrizable tvs E analytic if it contains a complete analytic vector subspace F such that E/F is analytic?~~

~~Also, the following question still seems to be open.~~

~~Problem 8.2 Does there exist a weakly analytic (DF )-space that is not analytic? Chapter 9 K-analytic and Analytic Spaces Cp(X)~~

Abstract This chapter deals with K-analytic and analytic spaces Cp(X). Some results due to Talagrand, Tkachuk, Velichko and Canela are presented. A remarkable theorem of Christensen stating that a metrizable and separable space X is σ -compact if and only if Cp(X) is analytic is proved. We show that the analyticity of Cp(X) for any X implies that X is σ -compact (Calbrix’s theorem). We show that Cp(X) X is K-analytic-framed in R if and only if Cp(X) admits a bounded resolution. We also gather several equivalent conditions for spaces Cp(X) to be Lindelöf spaces over locally compact groups X.

~~9.1 A theorem of Talagrand for spaces Cp(X)~~

~~We recall a couple of facts about Eberlein compact sets. Let X be a (Hausdorff) compact space. Then~~

~~X→ I Γ (9.1)~~

(i.e., X is embedded into a cube I Γ , where I =[−1, 1], and Γ denotes a set of continuous functions on X). A compact space X is called Eberlein compact [296] if this embedding can be chosen to take values in the Banach space c0 (Γ ) endowed with the pointwise convergence topology τp (see Theorem 19.29). The following result (see [27, Proposition IV.1.6], [149], [18]) characterizes Eberlein compact spaces. More results about Eberlein compact sets will be discussed in Chapter 19. Recall that a topological space X is called σ -compact if X is covered by a sequence of compact sets.

Theorem 9.1 The following assertions are equivalent for compact X: (i) X is Eberlein compact. (ii) There exists a compact space Y such that X is homeomorphic to a subspace of Cp(Y ). (iii) There exists a compact (resp. σ -compact) space Z ⊂ Cp(X) separating points of X. (iv) X is homeomorphic to a weakly compact subset in c0 (Γ ). (v) X is homeomorphic to a weakly compact subset of a Banach space.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 201 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_9, © Springer Science+Business Media, LLC 2011 202 9 K-analytic and Analytic Spaces Cp(X)

For the proof, we refer the reader to Section 19.17 or to the books [27] and [149]. In [318, Corollary 2.5], Okunev supplemented Theorem 12.6 by showing that: (o) If X is a σ -compact space and there exists a compact space Y such that X is homeomorphic to a subspace of Cp(Y ), the space Cp(X) is K-analytic. This fact will be used below. The following result is due to Arkhangel’skii [20] and Talagrand [388].

Proposition 9.1 (i) A compact space X is Eberlein compact if and only if Cp(X) has a dense σ -compact subspace. (ii) If X is Eberlein compact, then Cp(X) is an intersection of some countable family of σ -compact subspaces of RX.

~~A compact space X is called a Corson compact space if the embedding (9.1) can be chosen to take values in the Σ(Γ) space where~~

Γ Σ(Γ):= {x ∈ I :|{t ∈ I : x(t) = 0}| ≤ ℵ0}. Clearly, every compact metric space is Eberlein compact. The one-point compacti- ﬁcation of an uncountable discrete space provides a concrete Fréchet–Urysohn nonmetrizable Eberlein compact space; see [27, III.3.3]. A compact space X for which Cp (X) is K-analytic is called Talagrand compact. A compact space X is Gul’ko compact if Cp(X) is a Lindelöf Σ-space. Note that Eberlein compact ⇒ Talagrand compact ⇒ Gul’ko compact ⇒ Corson compact, and none of the reverse implication holds in general. Mercourakis [294, Theorems 3.1, 3.2, 3.3] characterized Talagrand (Gul’ko) compact spaces (see also Sokolov [385]). More facts concerning such compact sets will be discussed in Chap- ter 19. Talagrand, in his remarkable paper [388], proved the following theorem.

Theorem 9.2 (Talagrand) (i) If X is compact, then Cp(X) is K-analytic if and only if it has a compact resolution if and only if C(X) is weakly K-analytic. (ii) If X is compact and Cp(X) contains a K-analytic subspace separating the points of X, then Cp(X) is K-analytic.

Proof We prove only a part of (i). The remaining parts will be proved below in more general cases. Assume that Cp(X) is K-analytic. Then Cp(X) admits a compact N resolution {Kα : α ∈ N }.LetB be the closed unit ball in C(X). Note that {Kα ∩ B : α ∈ NN} is a compact resolution in the weak topology σ of C(X). Indeed, since Cp(X) is angelic by Theorem 4.5, it is enough to show that every sequence (fn)n in B that converges in Cp(X) converges in the weak topology of C(X). This is the case using the classical Lebesque dominated covergence theorem; see [213]. Since B in the topology σ is angelic, we apply Corollary 3.6 to get the K-analyticity of | = (B, σ B). Clearly, C(X) n nB, so the space C(X) is weakly K-analytic.

Tkachuk [399] extended the ﬁrst part of Theorem 9.2 (i) to any completely regular Hausdorff space X, and Canela [78] extended the other part of (i) for locally compact paracompact spaces X. 9.1 A theorem of Talagrand for spaces Cp(X) 203

~~Theorem 9.3 (Tkachuk) For a completely regular Hausdorff space X, the space Cp(X) is K-analytic if and only if Cp(X) has a compact resolution.~~

We extend Theorem 9.2 and Theorem 9.3. Recall that a (topological) space X is called hemicompact if it is covered by a sequence (Kn)n of compact subsets such that every compact subset of X is contained in some set Km. A space X is called a kR-space if every real-valued map deﬁned on X that is continuous on each compact subset of X is continuous.

Theorem 9.4 Let X be a hemicompact kR-space. If Cp(X) contains a subset S having a compact resolution and that separates points of X, the space Cp(X) has a compact resolution. Hence Cp(X) is K-analytic.

~~Proof Let Y be the algebra generated by S and the constant functions. Set~~

~~B = {x ∈ Y : |x (t)| ≤ 1,t ∈ X} .~~

~~N First assume that X is compact. Deﬁne the map ϕ : Y × B → Cp (X) by~~

~~∞ −n ϕ x,(xn)n = x + 2 xn. n=1~~

Since for each t ∈ X, p ∈ N, one has ∞ −n −p 2 xn(t) ≤ 2 , n=p+1 the map ϕ is continuous. By Proposition 3.10,thesetsY and B have a compact resolution, so ϕ Y × BN has a compact resolution as well. By the Stone–Weierstrass theorem, Y is uniformly dense in Cp (X). Hence N ϕ Y × B = Cp (X) .

This implies that Cp (X) has a compact resolution. Now assume that X is a hemicompact kR-space. Let (Kn)n be a fundamental (increasing) sequence of compact subsets of X. Note that for any n ∈ N the set Sn ={f |Kn : f ∈ S} has a compact resolution, belongs to Cp (Kn) and separates points of Kn. The ﬁrst part is applied ∞ to claim that each Cp (Kn) has a compact resolution. Hence n Cp (Kn) has the same property. On the other hand, since X is a hemicompact kR-space, the space ∞ Cp (X) is a closed subspace of n Cp (Kn),soCp (X) has a compact resolution, again by Proposition 3.10.

~~The assumptions in Theorem 9.4 are essential. If X is not hemicompact, the result fails. This follows from the following example. 204 9 K-analytic and Analytic Spaces Cp(X)~~

ℵ X Example 9.1 Let X be a discrete space of cardinality 2 0 . Then Cp(X) = R is not K-analytic and has a countable dense subset. Also, there exists a hemicompact space X that is not a kR-space, and Cp(X) is not K-analytic.

The first claim is clear by Theorem 7.1. For the other one, set X = N ∪{ξ} for ξ ∈ βN\N. Since every compact subset of X is finite, the space X is hemicompact. From Lutzer and McCoy (see [278]), the space Cp(X) is a Baire space. Moreover, Cp(X) is metrizable and not complete (since X is not discrete; therefore X is not a kR- space). The space Cp(X) is not K-analytic, otherwise Cp(X) would be a separable metrizable and complete space by Theorem 5.1 or Theorem 7.1. { : ∈ } By a σ -product of a family Xt t T of topological spaces with a basis at the ∈ point b t∈T Xt , we mean the subspace σb of t∈T Xt defined by

~~σb := x = (xt ) ∈ Xt :|t : xt = bt | < ∞ . (9.2) t∈T~~

If in the definition above finite is replaced by countable, σb is called a Σ-product at point b and is denoted by Σb. Recall that, according to Noble [315], every Σ- product is Fréchet–Urysohn provided each space Xt is first-countable. Clearly, if each Xt is a vector space, Σ0 is a vector subspace of t∈T Xt . We complete this section with the following result due to Okunev [318, Theo- rem 2.6].

~~Theorem 9.5 (Okunev) Let X be a σ -product of a family {Xt : t ∈ T } of Eberlein compact spaces. Then Cp(X) is K-analytic.~~

Proof It is easy to see that the σ -product of compact spaces is a σ -compact space. Since every space Xt is Eberlein compact, by Theorem 9.1 there exists a compact space Yt such that Xt is homeomorphic to a subspace of Cp(Yt ). We assume that the tth coordinate of the base point of the σ -product coincides with the zero func- := {∞} ∪ tion on Tt .LetY t∈T Yt be the one-point compactiﬁcation of the di- : → := rect sum t∈T Yt . Then the map j X Cp(Y ) deﬁned by j (x)(y) xt (y), y ∈ Yt , j(x)(∞) := 0 for all x ∈ X, is a homeomorphism onto the range. Now, applying Okunev’s result (o) mentioned above, we note that Cp(X) is K-analytic as claimed.

~~9.2 Theorems of Christensen and Calbrix for Cp(X)~~

All topological spaces in this section are completely regular and Hausdorff. If ξ ∈ βN \ N and X = N ∪{ξ}, the space Cp(X) is metrizable and not K-analytic, nevertheless, X is σ -compact; see Example 9.1. The following general question was posed by Christensen [99]. 9.2 Theorems of Christensen and Calbrix for Cp(X) 205

~~Problem 9.1 (Christensen) Let Cp(X) be an analytic space. Is X a σ -compact space?~~

A space X is called cosmic if X admits a countable network (i.e., a sequence (Un)n of subsets of X such that for each x ∈ X and every open subset G of X with x ∈ G there exists Un such that x ∈ Un ⊂ G. X is cosmic if and only if X is a continuous image of a metric separable space; see [17] and Proposition 3.5). By Proposition 6.4, X is analytic if and only if it is cosmic and K-analytic. Cosmic spaces look like a natural generalization of analyticity. Nevertheless, analytic spaces need not be metrizable, but any analytic space is always a cosmic space. We start with the following result due to Christensen [99].

~~Theorem 9.6 (Christensen) A metric and separable space X is σ -compact if and only if Cp(X) is analytic.~~

Proof Assume that Cp(X) is an analytic space. The idea of the proof of this part uses the argument from [30, Theorem 2.3]. Let (Y, d) be a compact metric space in which X is densely embedded. Then Z = Y \ X is separable and metrizable. First we show that Z is a Polish space. Set

R+ := {x ∈ R : x>0}. R R + := As there exists a (strictly increasing) homeomorphism of onto +, Cp (X) X Cp(X) ∩ R+ is an analytic subspace of Cp(X). Hence there exists a continuous map N φ : N → Cp(X) such that φ(NN) = C+(X). Since for each α ∈ NN the set {β ∈ NN : β ≤ α} is N N X compact in N , the function Γ : N → R+ deﬁned by

~~Γ(α):= inf{φ(β): β ≤ α},~~

~~N + for all α ∈ N , is well deﬁned. Note that Γ(β)≤ Γ(α)if α ≤ β, and iff ∈ C (X), ≥ ∈ NN { : ∈ NN} := \ then f Γ(α)for some α . Then Kα α , where Kα x∈X Y Bx and~~

Bx := {y ∈ Y : d(x, y) < Γ (α)(x)} is a compact resolution on Z that swallows compact sets in Z.LetA be compact in Z. Then the continuous map f : Y → R deﬁned by f(y):= d(y,A) restricted to X + belongs to C (X),sof |X ≥ Γ(α)for some α. Hence A ⊂ Kα. Now Theorem 6.1 is applied to deduce that Z is a Polish space. Therefore Z is a Gδ-subset of Y and hence Y \ Z = X is an Fσ -subspace of the space Y . Consequently, X is σ -compact and analytic. For the converse, we prove more than listed above. Let A ⊂ X be a dense subset of X.Let(Kn)n be an increasing sequence of compact sets in X covering X. We show, following Okunev’s argument from the proof of [319, Theorem 2.1] that 206 9 K-analytic and Analytic Spaces Cp(X) the space Cp(X|A) := {f ∈ Cp(A) : f admits a continuous extension over X} is analytic. Note that the restriction map from Cp(X) → Cp(X|A) is injective. Fix a metric d on X compatible with the topology of X. Set

~~A −1 Mkln := {f ∈[0, 1] :|f(x)− f(y)|≤k if~~

~~−1 −1 d(x,z)< n ,d(y,z)~~

For the converse inclusion, assume that f ∈ Cp(X|A,[0, 1]).Letf1 be a continuous extension of f to the space X.Nowﬁxk,l ∈ N. There exists n ∈ N such that −1 −1 −1 |f1(x) − f1(y)|≤k if d(x,z) < n and d(y,z) < n for some z ∈ Kl .This proves that f ∈ Mkln. A Next, we prove that each set Mkln is compact. Indeed, observe that [0, 1] \ A −1 Mkln is the set of all f ∈[0, 1] with |f(x)− f(y)| >k for some x ∈ A and −1 −1 A y ∈ A such that d(x,z) < n ,d(y,z)

~~Calbrix [77] answered Problem 9.1 by proving the following theorem.~~

~~Theorem 9.7 (Calbrix) If the space Cp(X) is analytic, then X is σ -compact. 9.2 Theorems of Christensen and Calbrix for Cp(X) 207~~

Theorem 9.7 will be deduced from some results presented below. It was extended by Arkhangel’ski and Calbrix [30]. Very recently, Ferrando and Kakol ˛ [160]showed the Arkhangel’skii–Cabrix theorem for spaces Cp(X) having a bounded resolution (see Proposition 9.6). The following example provides σ -compact spaces X such that Cp(X) is not Lindelöf; see [318, Example 2.7].

~~Example 9.2 (Okunev) There exists a topological space X that is a countable union of Eberlein compact spaces and for which Cp(X) is not Lindelöf.~~

Proof Let Z be a σ -product of uncountable copies of the compact space [0, 1] with zero base point. Let p be the point where all coordinates equal 1. Set X := Z ∪{p}. Clearly, X is the countable union of Eberlein compact sets. Now deﬁne a function f : X →[0, 1] by f(z)= 0 for each z ∈ Z and f(p)= 1. Since p does not belong to the closure of any countable subset of Z, the function f satisﬁes the following condition: (*) for any countable set A ⊂ X, the restriction f |A admits a continuous extension to the whole space X. Assume that Cp(X) is Lindelöf. Let A := {A ⊂ X countable}, and let

GA := {g ∈ C(X) : g|A = f |A} for each A ∈ A .LetD := {GA : A ∈ A }. Clearly, D has the countable intersection property. Since each GA is closed in Cp(X) , the condition (*) combined with the ∈ Lindelöf property of Cp(X) implies that A∈A GA is nonempty. Now choose h = A∈A GA. Then h f , a contradiction, since f is discontinuous.

~~We need the following two facts, the ﬁrst one from [285, Corollary 4.1.3], the other one from [27, Corollary IV.9.9].~~

~~Proposition 9.2 Cp(X) is cosmic if and only if X is cosmic.~~

~~Proposition 9.3 If X is a separable space that is not a cosmic space, υCp(X) is not a Lindelöf Σ-space.~~

A space X is called projectively σ -compact if every separable metrizable space Y that is a continuous image of X is σ -compact. Every σ -bounded space X (i.e., X is a countable union of topologically bounded subsets of X), is projectively σ -compact. Indeed, this follows from the fact that any σ -bounded paracompact topological space is σ -compact. The following result of Okunev [319] is essential in the proof of Theorem 9.7.

~~Proposition 9.4 A cosmic projectively σ -compact space X is σ -compact.~~

Proof Since a cosmic space that is projectively analytic is analytic (see [318, The- orem 1.3]), X is analytic, and then it is normal (see Lemma 6.1). Assume X is not 208 9 K-analytic and Analytic Spaces Cp(X)

σ -compact. Then it contains a homeomorphic copy F of NN as a closed subspace (see Corollary 3.13). There exists a homeomorphism r between the space F and the space of the irrational numbers. Let f : X → R be a continuous extension of r from X into R. Fix a continuous map g : X → R such that F = g−1(0). Deﬁne h : X → R2,h(x):= (f (x), g(x)), for all x ∈ X. Note that the set {(x, y) ∈ h(X) : y = 0} is closed in the range h(X) and is homeomorphic to the space NN. Since NN is not σ -compact, h(X) is not σ -compact, a contradiction.

~~The following theorem was obtained by Arkhangel’skii and Calbrix [30]; the proof uses the argument presented in the proof of Theorem 9.6.~~

~~X Theorem 9.8 (Arhhangel’skii–Calbrix) If Cp(X) is K-analytic-framed in R (i.e., X there exists a K-analytic space A such that Cp(X) ⊆ A ⊆ R ), then X is projectively σ -compact.~~

X Proof Assume Cp(X) is K-analytic-framed in R .LetY be a separable metrizable space, and let f : X → Y be a continuous map from X onto Y . We show that Y is σ - compact. The dual map f (g) := g ◦ f of RY → RX is a (linear) homeomorphism Y Y X of R onto the range f (Cp(Y )) ⊂ Cp(X), and f (R ) is closed in R .LetS be a X K-analytic subspace of R containing Cp(X). Then

~~ f (Cp(Y )) ⊂ Cp(X) ⊂ S, and S ∩ f (RY ) being closed in S is also K-analytic. Since~~

~~ Y f (Cp(Y )) ⊂ S ∩ f (R ),~~

~~Y we deduce that Cp(Y ) is K-analytic-framed in R . Set~~

~~+ := ∩ RY Cp (Y ) Cp(Y ) +.~~

+ It is easy to see (by using the exponential function) that Cp (Y ) is K-analytic-framed RY ⊂ RY + ⊂ in +. Therefore, there exists a K-analytic space T + with Cp (X) T and an usco compact-valued map φ from NN covering T . We complete the proof in a similar manner as in the proof of Theorem 9.6.

As usual, for a topological space X,byβX and υX we denote the Stone–Cechˇ compactiﬁcation and the realcompactiﬁcation of X, respectively. By f υ : υX → R, we denote the unique extension of f ∈ C(X). We need the following simple lemma.

υ Lemma 9.1 (a) The map Φ : Cp(X) → Cp(υX) deﬁned by f → f is linear, injective and surjective. The converse map Φ−1 is continuous. ⊂ ∈ ∈ υ = (b) If (gn) Cp(X) and y υX, there exists x X such that gn (y) gn(x) for every n ∈ N. (c) If a subset A of Cp(X) is countable, the map Φ|A is continuous. 9.2 Theorems of Christensen and Calbrix for Cp(X) 209

Proof (a)isclear. := − υ ∈ ∈ N (b) Set fn(x) gn(x) gn (y) for all x X and n . Then each fn is contin- υ = ∈ N uous, and fn (y) 0 for all n . Set −n g(x) := min{2 , |fn(x)|} n for all x ∈ X. Clearly, g ∈ C(X) and gυ (y) = 0. Assume that g(x) = 0 for each x ∈ X.Letz(x) := g−1(x) for all x ∈ X. Then zυ (x)gυ (x) = 1 for each x ∈ X. Hence zυ (y)gυ (y) = 1 by the density of X in υX, a contradiction. Hence the conclusion holds. (c) Assume (fγ ) ⊂ A,f0 ∈ A and fγ → f0 in Cp(X). Let y ∈ υX. By (b) there υ is x ∈ X with f(x)= f (y) for every f ∈ A. Then fγ (x) → f0(x). Hence υ → υ fγ (y) f (y). υ → υ | Consequently, fγ f in Cp(υX), so the map Φ A is continuous.

~~The following lemma can be found in [83].~~

~~Lemma 9.2 The space Cp(X) is angelic if and only if Cp(υX) is angelic.~~

Proof If Cp(X) is angelic, Cp(υX) is angelic by Lemma 9.1 and Theorem 4.1 (angelic lemma). To prove the converse, assume that Cp(υX) is angelic. If A ⊂ υ υ C(X) is a relatively countably compact set in Cp(X),thesetA := {f : f ∈ A} is relatively countably compact in Cp(υX) because (by Lemma 9.1) the restriction map is a homeomorphism on any countable subset of Cp(υX). By hypothesis, each relatively countably compact set A in Cp(υX) is relatively compact, and the closure υ of A equals the sequential closure. The same conclusion holds for A in Cp(X), which proves that Cp(X) is angelic.

~~For Proposition 9.6, we need the following result due to Arkhangell’ski (i) and Okunev (ii); see [27, Proposition IV.9.4] or [318], [26].~~

Proposition 9.5 (i) If υX is a Lindelöf Σ-space, there exists a Lindelöf Σ-space X Z such that Cp(X) ⊂ Z ⊂ R . X (ii) If there exists a Lindelöf Σ-space Z such that Cp(X) ⊂ Z ⊂ R , υX is a Lindelöf Σ-space.

There exists a Lindelöf space X that is not K-analytic and for which Cp(X) is K- analytic; see [318], [267]. For such X, by Proposition 9.5 (ii), the space υX is a Lindelöf Σ-space. This follows also from the following fact: If Cp(X) is a Lindelöf Σ-space, υX is a Lindelöf Σ-space. Indeed, since Cp(X) is a Lindelöf Σ-space, applying Proposition 9.5 (i), we note that there exists a Lindelöf Σ-space Z such that

~~Cp(X) Cp(Cp(X)) ⊂ Z ⊂ R . 210 9 K-analytic and Analytic Spaces Cp(X)~~

Clearly, X is included in Cp(Cp(X)).LetY be the closure of X in Z. Then Y is a Lindelöf Σ-space. Since every continuous map on X can be extended to a continuous map over RCp(X), every continuous map on X can be extended to a continuous map to Y . Hence υX = Y is a Lindelöf Σ-space. Proposition 9.6 was proved in [160].

Proposition 9.6 (Ferrando–Kakol) ˛ The following conditions are equivalent: (i) Cp(X) admits a bounded resolution. X (ii) Cp(X) is both K-analytic-framed in R and angelic. X (iii) Cp(X) is K-analytic-framed in R . (iv) For any tvs Y containing Cp(X), there exists a space Z such that Cp(X) ⊆ Z ⊆ Y and Z admits a resolution consisting of Y -bounded sets. ⇒ : ∈ NN Proof (i) (ii):Let Aα α be a bounded resolution for Cp(X).ByBα X N denote the closure of Aα in R . Set Y = Bα : α ∈ N . Note that each Bα is a compact subset of RX and X Cp (X) ⊆ Y ⊆ R . As usual, set T (α | n) = Bβ : β (i) ≤ α (i) , 1 ≤ i ≤ n N N for each (α, n) ∈ N × N and Cα = {T (α | n) : n ∈ N} for each α ∈ N . Then N {Cα : α ∈ N } covers Y , Bα ⊆ Cα, Cα ⊆ Cβ if α ≤ β, and each Cα is countably compact. Deﬁne N S = Cα : α ∈ N and put N (α | n) = T (α | n) (the closure in RX). Note that

~~N N (α | n) : (α, n) ∈ N × N~~

~~N is a countable network modulo the family {Cα : α ∈ N } of compact subsets of S. Indeed, we prove that for a neighborhood U of Cα in S, there is m ∈ N such that~~

~~Cα ⊆ N (α | m) ⊆ U.~~

~~There is an open neighborhood V of Cα such that~~

~~Cα ⊆ V ⊆ V ⊆ U.~~

Note also that there is m ∈ N such that T (α | m) ⊆ V ; otherwise there exist a se- N quence (βn)n in N with βn (i) ≤ α (i) for 1 ≤ i ≤ n and a sequence (yn)n in S \ V ∈ ∈ N = { : ∈ N} ≤ with yn Bβn for each n . Set δ (i) max βn (i) n . Then βn δ. Hence the sequence (yn)n has a cluster point y ∈ Bδ \ V . Since δ ≤ α,wehave

y ∈ Bδ ⊆ Cδ ⊆ Cα ⊆ V, 9.2 Theorems of Christensen and Calbrix for Cp(X) 211 which yields a contradiction. From Cα ⊂ T(α|m) ⊂ V , it follows that Cα ⊂ N(α|m) ⊂ U. This shows that S is a Lindelöf Σ-space by Proposition 3.5. Hence S is a Lindelöf N space. Since Cα : α ∈ N is a compact resolution for S, the space S is a quasi- Suslin space. As every Lindelöf quasi-Suslin space is K-analytic, S is K-analytic. Clearly, X Cp(X) ⊂ S ⊂ R . X This shows that Cp(X) is K-analytic-framed in R . By Proposition 9.5, we note that υX is a Lindelöf Σ-space. Finally, as each Lindelöf Σ-space is web-compact by Example 4.1 (2), we apply Theorem 4.5 to deduce that Cp(υX) is angelic. Hence Cp(X) is angelic by Lemma 9.2. (ii) ⇒ (iii) is clear. N (iii) ⇒ (iv): If L is a space having a compact resolution Aα : α ∈ N , and

X Cp(X) ⊆ L ⊆ R , then N Aα ∩ Cp(X) : α ∈ N is a bounded resolution in Z := Cp(X) consisting of bounded sets in any tvs Y topologically containing Cp(X). The implication (iv) ⇒ (i)isobvious.

~~The equivalence between conditions (i), (iii) and (iv) below was proved in [30, Theorem 2.4].~~

Corollary 9.1 Let X be a cosmic space. The following assertions are equivalent: (i) X is σ -compact. (ii) Cp(X) has a bounded resolution. X (iii) Cp(X) is K-analytic-framed in R . X (iv) Cp(X) is analytic-framed in R .

Proof (i) ⇒(ii): Note that if X is σ -compact, Cp(X) has a bounded resolution. N Indeed, if X is covered by a sequence (Cn)n of compact sets, then Aα : α ∈ N with Aα = f ∈ C (X) : sup |f (x)| ≤ an,n∈ N , x∈Cn N where α = (an) ∈ N is a bounded resolution for Cp (X). (ii) ⇒ (iii): this follows from Proposition 9.6. (iv) ⇒ (iii): Is obvious. (iii) ⇒ (i): This follows from Theorem 9.8 and Proposition 9.4. (i) ⇒ (iv): Since X is cosmic and covered by a sequence (Kn)n of compact := sets, the topological sum Y n Kn provides a metrizable (recall that a cosmic compact space is metrizable) separable σ -compact space. By Theorem 9.6, 212 9 K-analytic and Analytic Spaces Cp(X) the space Cp(Y ) is analytic. If f : Y → X is a canonical map from Y onto X, its dual map f (g) := g ◦ f of RX → RY is a homeomorphism of RX onto the range X Y f (Cp(X)) ⊂ Cp(Y ), and f (R ) is closed in R . Hence

~~ X f (Cp(X)) ⊂ f (R ) ∩ Cp(Y ),~~

~~ X and f (R ) ∩ Cp(Y ) is analytic (being closed in Cp(Y )).~~

~~The following corollary extends Corollary 7.2.~~

~~Corollary 9.2 For a metric space X, the following assertions are equivalent: (i) X is σ -compact. (ii) Cp(X) has a bounded resolution. (iii) Cp(X) is analytic.~~

Proof (i) ⇒ (ii) was proved in Corollary 9.1. (ii) ⇒ (iii): Since Cp(X) has a bounded resolution, by Proposition 9.6 the space X Cp(X) is K-analytic-framed in R and Cp(X) is angelic. A metric space X for which Cp(X) is angelic is separable by Corollary 6.10. Hence X is a cosmic space and Corollary 9.1 applies to get that X is σ -compact. Now Theorem 9.6 is applied to deduce that Cp(X) is analytic. (iii) ⇒ (i) by Theorem 9.7.

~~From Corollary 9.1, we have the following corollary.~~

~~N Corollary 9.3 The space Cp(N ) does not admit a bounded resolution.~~

The following observation is due to Christensen [99]: If X is a kR-space, there is N no continuous surjection from X onto Cc(X). Consequently, the space Cc(N ) is not analytic. Indeed, assume that there is a continuous surjection φ from X onto Cc(X). Deﬁne a real-valued function f by f(x)= 1 + φ(x)(x) for all x ∈ X. Since X is a kR-space and f is continuous on each compact subset of X,themapf is continuous. Fix x ∈ X such that φ(x)= f . Then 1 + φ(x)(x) = f(x)= φ(x)(x), a contradiction.

~~Corollary 9.4 If Cp(X) has a bounded resolution, a compact set A ⊂ Cp(X) is metrizable if and only if A is contained in a separable subset of Cp(X).~~

Proof Assume a compact set A ⊂ Cp(X) is contained in some separable subset υ υ υ of Cp(X). Set A := {f : f ∈ A}. Then A is contained in a separable subset of Cp(υX) (see Lemma 9.1(c)). Since υX is web-compact (the proof of Proposi- tion 9.6 (i) ⇒ (ii) showed that υX is a Lindelöf Σ-space), we apply Corollary 4.5 to deduce that Aυ is metrizable. Since Aυ and A are homeomorphic, the conclusion follows. 9.2 Theorems of Christensen and Calbrix for Cp(X) 213

~~Now we are ready to prove Theorem 9.7 as a simple consequence of results above.~~

Proof If Cp(X) is analytic, then Cp(X) is a cosmic space. Hence X is cosmic by Proposition 9.2. Next, by Proposition 9.6, the space Cp(X) is K-analytic-framed in RX. This, together with Theorem 9.8, implies that X is projectively σ -compact. Proposition 9.4 yields that X is σ -compact.

~~Theorem 9.9 extends Theorem 9.3.~~

Theorem 9.9 Let ξ be a regular topology on C(X) stronger than the pointwise one. The following assertions are equivalent: (i) (C(X), ξ) is K-analytic. (ii) (C(X), ξ) is quasi-Suslin. (iii) (C(X), ξ) admits a relatively countably compact resolution.

~~Proof Since, by Proposition 9.6, each condition listed above implies the angelicity of Cp(X), Theorem 4.1 yields the angelicity of (C(X), ξ), and then Corollary 3.6 applies.~~

Theorem 9.9 fails in general for spaces of the form Cp(X, Y ) if Y is an arbitrary metric space. Let Y := [−1, 1] be endowed with the natural topology from R.The following example is due to Tkachuk [399].

~~Example 9.3 Let X be the Lindelöﬁcation of the discrete space ω1. Under the CH, the space Cp(X, [−1, 1]) is not Lindelöf and admits a compact resolution.~~

Proof Set I := [−1, 1] and X := ω1 ∪{x}, where ω1 consists of isolated points, and Ux is an open neighborhood of x in X if |X \ Ux| is countable. Since Cp(X, I) is countably compact and not compact [398], the space Cp(X, I) is not Lindelöf. Note also that Cp(X, I) = Y × I, where

~~Y := {f ∈ C(X,I) : f(x)= 0}~~

~~X ={f ∈ I : f(x)= 0 ∧∃t<ω1 : f(r)= 0 ∀ r ≥ t}. By Proposition 3.10, it is enough to show that Y has a compact resolution. Assume the CH. Then Y ={ft : t<ω1}, and there exists a set~~

~~N Z := {αt : t<ω1}⊂N~~

~~∗ N ∗ N such that αt ≤ αr if t~~

Kα := {ft : t ∈ Nα} is relatively compact in Y since the closure of any countable subset of Y is compact N and Kα ⊂ Kβ if α ≤ β for α, β ∈ N .Iff ∈ Y , there exists t<ω1 such that f = ∈ ft Kαt .

~~The space Cp(X, I) is not angelic. Note that the space Cp(X) for X from Exam- ple 9.3 is not K-analytic. This can also be deduced from the following general fact extending Theorem 9.9.~~

Proposition 9.7 Assume that Cp(X) is K-analytic. Let Z be a metric space. Let ξ be a regular topology on the space C(X,Z) that is stronger than the pointwise topology of C(X,Z). The following assertions are equivalent: (i) (C(X, Z), ξ) is K-analytic. (ii) (C(X, Z), ξ) admits a compact resolution. (iii) (C(X, Z), ξ) admits a relatively countably compact resolution.

~~Proof Since Cp(X) is K-analytic, by Proposition 9.6 the space Cp(X) is angelic. By Theorem 4.3, the space Cp(X, Z) is also angelic. Now we proceed as in Theo- rem 9.9.~~

~~Corollary 9.5 is from [83].~~

~~Corollary 9.5 Let X be a web-compact space. Then Cp(X) is K-analytic if and only if Cp(υX) is K-analytic.~~

Proof By Theorem 4.5, the space Cp(X) is angelic. Assume that Cp(υX) is K- analytic. Since the restriction map from Cp(υX) onto Cp(X) is a continuous bijection, Cp(X) is K-analytic. Assume Cp(X) is a K-analytic space. Then Cp(X) N admits a compact resolution {Kα : α ∈ N }. { υ : ∈ NN} Since Cp(υX) is angelic by Lemma 9.2, the following family Kα α is υ := { υ : ∈ } a compact resolution in Cp(υX), where Kα f f Kα . Indeed, every set υ Kα is relatively countably compact; see the proof of Lemma 9.2. Since Cp(υX) υ ∈ NN is angelic, the set Kα is relatively compact for each α . Now Theorem 9.9 is aplied applies to deduce that the space Cp(υX) is K-analytic.

~~We conclude this section with the following application of Theorem 9.7.~~

~~Proposition 9.8 Cp(X) is an analytic space if and only if it admits a stronger metrizable and analytic vector topology ξ such that ξ is polar with respect to the topology of Cp(X).~~

Proof Assume that Cp(X) is analytic. The spaces Cp(X) and Cc(X) are separable; see [213, Theorem 2.10.3]. By Theorem 9.7, the space X is σ -compact. Let (Kn)n 9.3 Bounded resolutions for Cp(X) 215 be an increasing sequence of compact sets in X covering X. For each k ∈ N,set −n −k Vk := f ∈ C(X) : 2 min 1, sup |f(x)| ≤ 2 . ∈ n x Kn It is easy to see that

Vk+1 + Vk+1 ⊂ Vk, and Vk is balanced and absorbing for each k ∈ N.Letτc and τp denote the original topologies of the spaces Cc(X) and Cp(X), respectively. Let ξ be a vector topology on C(X) whose basis of neighborhoods of zero is formed by the sets Vk, k ∈ N. Clearly, ξ ≤ τc. Note also that τp ≤ ξ. Indeed, let y ∈ X,0<ε<1, and set

U := {f ∈ C(X) :|f(y)| <ε}. ∈ N ∈ Then U is a τp -neighborhood of zero. There exists m such that y Km.One −1 −p proves that Vp ⊂ ε 2 U. This shows our conjecture. Therefore ξ is a metrizable and separable vector topology stronger than τp. Moreover, ξ admits a basis of neighborhoods of zero consisting of τp-closed sets. By Corollary 6.13, we obtain that ξ admits a complete resolution. Corollary 6.14 proves that (C(X), ξ) is analytic. The converse implication is clear.

It still seems to be unknown if the K-analyticity of Cp(X) implies that X is σ -bounded [30]. Finally, note that if X is a σ -compact space, then X × Y is Lin- delöf for each Lindelöf space Y . What about the converse implication? Alster [3, Theorem] provided under Martin’s axiom an interesting sufﬁcient condition for a metrizable analytic space E to be σ -compact; for the proof, we refer to [3].

~~Theorem 9.10 Under Martin’s axiom, if X is metrizable and analytic and such that the product X with every Lindelöf space is Lindelöf, then X is σ -compact.~~

~~An interesting approach (in terms of the reminder βX \ X) to determine if a topological group X is σ -compact was provided in [33] and [32].~~

~~9.3 Bounded resolutions for Cp(X)~~

A topological space Y has tightness m (we denote t(Y ) ≤ m) if for each set A ⊂ Y and every x ∈ A there exists a subset B ⊂ A of cardinality m such that x ∈ B.Ifthis holds for m =ℵ0, we say, as usual, that Y has countable tightness. Recall [339] and seealso[27, Theorem II.1.1] for the following proposition.

n Proposition 9.9 (Arkhangel’skii–Pytkeev) t(Cp(X)) ≤ m if and only if (X ) ≤ m for each n ∈ N, where (X) denotes the Lindelöf number of X. Hence, if X is a Lindelöf Σ-space, the space Cp(X) has countable tightness. 216 9 K-analytic and Analytic Spaces Cp(X)

~~We recall also the following result due to Asanov [35]; see also [27, Theorem I.4.1].~~

~~Theorem 9.11 (Asanov) For every completely regular Hausdorff space X, we have n t(X ) ≤ (Cp(X)) for each n ∈ N.~~

Hence, if Cp(X) is Lindelöf, every ﬁnite product of X has countable tightness. A family U of nonempty open subsets of a topological space X is called a π- base if for every nonempty open set V in X there exists U ∈ U such that U ⊂ V . A family U of nonempty open subsets in X is called a local π-base at x if for each neighborhood V of x there exists U ∈ U such that U ⊂ V . Set

~~πχ(x,X):= min {|U |:U is a local π-base of x}+ℵ0. := Let πχ(X) supx∈X πχ(x, X). The following result of Shapirovskii [375] has numerous consequences.~~

Proposition 9.10 If X is a compact space such that t(X) ≤ m, the space X has a π-base such that πχ(X) ≤ m. Consequently, every compact space with countable tightness has a countable local π-base at each point x.

~~Proposition 9.10 is used to prove the following proposition [401, Proposition 2.1]; seealso[402] and [396].~~

~~Proposition 9.11 A topological group G that embeds (as a topological space) into a compact space K of countable tightness is metrizable.~~

Proof We may assume that G is dense in K and t(K) ≤ℵ0. By Proposition 9.10, we have πχ(K)≤ℵ0. Then πχ(x,G)= πχ(x,K)≤ℵ0 for each x ∈ G. Since G is a topological group, χ(x,G) = πχ(x,G) ≤ℵ0 for any x ∈ G, where χ(x,G) denotes the character of x; see, for example, [34, Proposition 5.2.6]. This clearly implies that G is metrizable.

~~To prove Theorem 9.12, we need three lemmas (see [401]).~~

Lemma 9.3 Let (Fn)n be an increasing sequence of subsets of X. Let y ∈ \ { ∈ : X n Fn, and let Z be a compact subset of Cp(X) contained in f Cp(X) f(y)= 0}. If there exists ε>0 such that for each n ∈ N there exists fn ∈ Z such | | ∈ ∈ that fn(x) ε for each x Fn, then y/ n Fn. | | Proof Let f be an adherent point of the sequence (fn)n. Then f(x) ε for each ∈ ∈ | | = x n Fn. By the continuity of f ,ifz n Fn, then f(z) ε. Since f(y) 0, ∈ we conclude that y/ n Fn.

Lemma 9.4 Let(Fn)n be a sequence of closed subsets of X. Assume Cp(X) is σ -compact. Then n Fn is a closed subset of X. Hence, if Cp(X) is σ -compact, the space X is a P -space (i.e., every Gδ-set in X is open). 9.3 Bounded resolutions for Cp(X) 217

~~Proof Without loss of generality, we may assume that (Fn)n is increasing. Choose y ∈ X\ Fn. n~~

~~Let (Wn)n be an increasing sequence of compact sets covering Cp(X).IfZm := {f ∈ Wm : f(y)= 0}, then {f ∈ Cp(X) : f(y)= 0}= Zm. (9.3) m~~

Assume that for each m ∈ N the set Zm does not verify Lemma 9.3. Then, for = −m ∈ ∈ ε 2 , we ﬁnd nm such that for each f Zm there exists yf,m Fnm such that −m f(yf,m) < 2 . (9.4) := : →[ −m] Let Cm Fnm , and let gm X 0, 2 be a continuous function such that −m gm(y) = 0 and gm(Cm) ={2 }. Deﬁne a continuous function g : X →[0, 1] by g(x) := Σ{gn(x) : n ∈ N}. Then we have

~~g(y) = 0. (9.5)~~

~~Note that g ∈ Zm for each m ∈ N. Indeed, otherwise~~

~~−m g(yg,m) gm(yg,m) = 2 . (9.6)~~

Therefore g/∈ Zm. (9.7) m This ﬁnally yields a contradiction. Therefore there exists msuch that Z := Zm veri- = −m ∈ ﬁes Lemma 9.3 for ε 2 . Then y/ n Fn, proving that n Fn is a closed subset of X.

~~Lemma 9.5 Let {xn : n ∈ N} be an inﬁnite set in a P-space X. Then, for a sequence (αn)n of real numbers, there exists a continuous function f : X → R such that f(xn) = αn for n ∈ N.~~

Proof Clearly, all subsets of {xn : n ∈ N} are closed. Therefore, by an easy induction, there exists a sequence of open sets (Un)n such that each set Un is an open neighborhood of xn and {Un : n ∈ N} is a family of pairwise disjoint subsets of X. { : ∈ N} ∈ The family Un n is discrete. Indeed, if z/ n Un, then W := X\ Un n ∩ =∅ ∈ N ∈ is an open neighborhood of z such that W Un for n . Moreover, if z Up, ∈ := \ then z/ n=p Un and W X n=p Un is an open neighborhood of z such that 218 9 K-analytic and Analytic Spaces Cp(X)

W ∩ Un =∅for n ∈ N\{p}. Since for each n ∈ N there exists a continuous function ϕn : X → R such that ϕn(xn) = αn, and ϕn(X\Un) ={0}, |ϕn(x)| ≤ |αn|, for each x ∈ X, the continuous function f : X → R, deﬁned by f(x):= ϕn(x), n satisﬁes f(xn) = αn for each n ∈ N.

~~We are ready to prove Theorem 9.12;see[27, Theorem I.2.1]. Recall that a subset F of X is topologically bounded if f(F)is bounded for each f ∈ Cp(X).~~

~~Theorem 9.12 (Velichko) The space Cp(X) is σ -compact if and only if X is ﬁnite.~~

Proof Let (Wn)n be an increasing sequence of compact subsets of Cp(X) covering Cp(X). By Lemma 9.5 and Lemma 9.4, every topologically bounded subset of X is ﬁnite. By Proposition 2.17, the space Cp(X) is barrelled. Since the absolutely convex closed envelope Bn of each Wn is a bounded set, we apply Proposition 2.13 to deduce that X R = Cp(X) = Bn = Bn, n n where the closure is taken in RX. By the Baire category theorem, there exists m ∈ N X such that Bm is a neighborhood of zero in R , and hence X is ﬁnite.

Theorem 9.12 was extended by Tkachuk and Shakhmatov [403]forσ -countably compact spaces Cp(X) (i.e., Cp(X) is covered by a sequence of countably compact sets; see also [27, Theorem I.2.2]). Note, however, that according to Proposi- tion 9.6,aCp(X) having a bounded resolution is angelic. Since for angelic spaces (relatively) countable compact sets and (relatively) compact sets are the same, we see that Tkachuk and Shakhmatov’s result follows from Velichko’s theorem. In fact, we have the following theorem.

~~Theorem 9.13 The following assertions are equivalent for X: (i) X is ﬁnite. (ii) Cp(X) is σ -compact. (iii) Cp(X) is σ -relatively countably compact. (iv) Cp(X) is σ -countably compact.~~

A similar result was obtained in [398]forσ -bounded spaces Cp(X) (i.e., Cp(X) is covered by a sequence of topologically bounded sets). The proof of Proposition 9.12 uses Corollary 9.3 and the argument from [30, Proposition 3.1].

Proposition 9.12 (Tkachuk) If Cp(X) is σ -bounded, the space X is pseudocompact and every countable set in X is discrete. Consequently, every compact set in X is ﬁnite. 9.3 Bounded resolutions for Cp(X) 219

Proof Assume X is not pseudocompact. Hence X contains a copy of the discrete space N, also named N, and the continuous map T : Cp(X) → Cp(N) deﬁned by T(f):= f |N is surjective. Therefore Cp(N) is σ -bounded, so Cp(Cp(N)) has a N N bounded resolution. Since Cp(N ) ⊂ Cp(Cp(N)), it follows that Cp(N ) has a bounded resolution, a contradiction with Corollary 9.3. Hence X is pseudocompact. Now we prove that every countable set in X is discrete. Let D ⊂ X be a countable subset of X.ByCp(D|X) we denote the subspace of Cp(D) that is the image of Cp(X) under the restriction map. As Cp(D) is separable and metrizable, Cp(D|X) is a cosmic space. Since Cp(D) is σ -bounded, the space Cp(D|X) is σ -bounded, and then Cp(Cp(D|X)) has a bounded resolution. By Corollary 9.1, Cp(D|X) is σ -compact. Using a similar argument as in the proof of Lemma 9.4, we deduce that D is a P-space. Since it is countable, it must be discrete.

Proposition 9.12 shows that a realcompact space X is ﬁnite if and only if Cp(X) is σ -bounded. Indeed, since every realcompact pseudocompact space is compact, we may apply Proposition 9.12. Note that, by [25, Proposition 9.31] (see also [30, Remark]), there exists an inﬁnite space X such that Cp(X) is σ -bounded.

~~Corollary 9.6 Let Cp(X) be a σ -bounded space. Then X is countable if and only if X is a K-analytic space.~~

N Proof If X is K-analytic, the space X admits a compact resolution {Kα : α ∈ N }. = Since, by Proposition 9.12, every compact set in X is ﬁnite, the space X α Kα is countable by Proposition 3.7. The converse is obvious.

Using Theorem 9.7 and Theorem 9.12, we have that if Cp(Cp(X)) is analytic, then X is ﬁnite. Indeed, the analyticity of Cp(Cp(X)) implies that Cp(X) is σ - compact, and then Theorem 9.12 applies. We have even more: If Cp(Cp(X)) is K- analytic, the space X is ﬁnite by [27, IV.9.21]. We note also the following corollary.

Corollary 9.7 For a realcompact space X, the space Cp(Cp (X)) has a bounded resolution if and only if X is finite. Hence, if Cp(Cp(X)) is K-analytic, the space X is finite. Proof Assume that Cp Cp (X) has a bounded resolution. Then, by Corollary 9.1, C(X) the space Cp Cp (X) is K-analytic-framed in R . Hence there is a K-analytic space Y such that C(X) Cp(Cp (X)) ⊆ Y ⊆ R . Then every compact subset of X is finite (see [30, Corollary 3.4]). Since

X ⊆ Y ⊆ RC(X) and X is realcompact, X is a closed subspace of Y . Hence X is a K-analytic space whose compact sets are ﬁnite. Thus X is countable by Proposition 3.7. Conse- quently, Cp (X) is a separable metric space and hence a cosmic space. Applying 220 9 K-analytic and Analytic Spaces Cp(X)

~~Corollary 9.1, we derive thatCp (X)is σ -compact. By Theorem 9.12, X is ﬁnite. Conversely, if X is ﬁnite, Cp Cp (X) has a bounded resolution.~~

~~Corollary 9.8 provides another Velichko-type result; this follows also from Proposition 9.12.~~

Corollary 9.8 A realcompact space X is ﬁnite if and only if Cp(X) is σ -bounded. Proof If Cp(X) is covered by a sequence of topologically bounded sets, Cp Cp (X) has a bounded resolution; see the proof of Corollary 9.1,(i)⇒ (ii). Now Corol- lary 9.7 applies.

It is known that Cp(X) is a Fréchet space if and only if X is countable and discrete. Applying Theorem 6.1, we characterize Fréchet spaces Cp(X) as spaces having a special resolution. We need the following two results; the ﬁrst one is due to Agryros and Negrepontis [14](seealso[24, Theorem 7.25]) and the other one follows from [399].

~~Proposition 9.13 If X is a Gul’ko compact space, the density, weight and Suslin number of X coincide.~~

~~Proposition 9.14 If Cp(X) has a compact resolution swallowing compact sets, the same property holds for Cp(υX).~~

Proof Let T : Cp(υX) → Cp(X) be the restriction map from Cp(υX) onto Cp(X). −1 It is enough to show that for every compact set K ⊂ Cp(X) the set T (K) is −1 compact in Cp(υX).LetA ⊂ T (K) be a countable subset. T |A : A → T(A)is a homeomorphism by Lemma 9.1. Therefore A has a cluster point in W := T −1(K), so W is countably compact. By Proposition 9.6, the space Cp(X) is angelic. Since, by Lemma 9.2, the space Cp(υX) is angelic (in angelic spaces, countably compact sets are compact), we obtain that W is compact.

~~Now we prove the following result [399, Theorem 3.7].~~

~~Theorem 9.14 (Tkachuk) If Cp(X) admits a compact resolution swallowing compact sets, X is countable and discrete.~~

Proof Since Cp(X) has a compact resolution, it is K-analytic, and hence X ⊂ Cp(Cp(X)) is angelic by Theorem 4.5. We claim that every compact set in X is ﬁnite. Assume that K is an inﬁnite compact subset of X.AsX is angelic, there exists a sequence {xn : n ∈ N} of different elements in K that converges to x ∈ K. Since every compact countable subset of K is metrizable by Proposition 9.13,the compact set K0 := {xn : n ∈ N}∪{x} is metrizable. There exists a continuous linear extender T : Cp(K0) → Cp(X) such that T(f)|K0 = f for each f ∈ C(K0) (see [24, Proposition 4.1], or [146], [29], [138], [372]ifX is metrizable). Hence 9.3 Bounded resolutions for Cp(X) 221

Cp(K0) embeds in Cp(X) as a closed subset. Consequently, Cp(K0) admits a compact resolution swallowing compact sets. Now Theorem 6.1 is applied to deduce that Cp(K0) is a Polish space, and then K0 is discrete and hence finite, a contradiction. We proved that every compact set in X is finite. Since, by Proposition 9.14, the space Cp(υX) admits a compact resolution swallowing compact sets, every compact set in υX is finite. As υX is a Lindelöf Σ-space by Proposition 9.5, we use Corollary 3.4 to get that υX is countable. Hence Cp(X) is a separable and metrizable lcs admitting a compact resolution swallowing compact sets. Theorem 6.1 is applied to deduce that Cp(X) is a Polish space, and hence X is countable and discrete; see [27, Corollary I.3.3].

The proof above uses some extensions of the Tietze–Urysohn theorem. We refer the reader to articles [142], [202] and [201] for more information. For example, in [202, Corollary J] it is shown that there exist compact separable spaces X having closed subspaces Y that contain uncountable disjoint collections of relatively open sets and no continuous extenders from Cp(Y ) into Cp(X). An lcs E is called web-bounded if it admits a family {Cα : α ∈ Σ} of subsets of E N covering E for some nonempty subset Σ ⊂ N such that, if α = (nk) ∈ Σ and xk ∈ ⊂ Cn1,n2,...,nk , the sequence (xk)k is bounded in E. Since Cα Cn1,n2,...,nk , each set Cα is bounded (so the family {Cα : α ∈ Σ} is bounded). Indeed, it is enough to see ∈ N ⊂ that for each neighborhood of zero U in E there exists k such that Cn1,n2,...,nk kU. A topological space X is called almost quasi-Suslin if there exists a nonempty N set Ω ⊂ N and a map T from Ω into subsets of E such that, if αk → α in Ω and xk ∈ T(αk), then (xk)k has an adherent point belonging to T(α). If the same holds for Ω = NN, the space X is already known as a quasi-Suslin space. The following simple observation can be found in [152].

~~Proposition 9.15 If X is almost quasi-Suslin (quasi-Suslin), the space υX is Lin- delöf Σ (K-analytic). Hence X is a Lindelöf Σ-space if and only if X is Lindelöf and almost quasi-Suslin.~~

Proof We prove only the case where X is an almost quasi-Suslin space. Since every T(α) is countably compact, its closure T(α) in υX is compact. It is easy to see → := that the map α T(α)is usco, so Z α∈Σ T(α)is a Lindelöf Σ-space. Since X ⊂ Z ⊂ υX, the space Z = υZ = υX is a Lindelöf Σ-space. A topological space X is called web-bounding if X = {Aα : α ∈ Σ} for some ⊂ NN = ∈ ∈ ∈ N nonempty subset Σ and, if α (nk) Σ and xk Cn1,n2,...,nk for all k , the set {xk : k ∈ N} is topologically bounded; see [320]. If X = {Aα : α ∈ Σ},X will be called strongly web-bounding. We need the following result of Nagami [308] that supplements Proposition 3.5 above; see also [27, Proposition IV.9.2] or [402, Theorem 2.1].

Proposition 9.16 A topological space X is a Lindelöf Σ-space if and only if in some (hence in any) compactiﬁcation bX of X there is a countable family F of 222 9 K-analytic and Analytic Spaces Cp(X) compact sets such that, if x ∈ X and y ∈ bX \ X, there exists B ∈ F for which x ∈ B and y/∈ B.

Blasco [59] proved that for a separable space X its realcompactiﬁcation υX is a Lindelöf space if and only if every base in X is complete. Spaces X for which υX is a Lindelöf space were called pseudo-Lindelöf ; see also [208]. Next, Theorem 9.15 characterizes those spaces X for which the realcompacti- ﬁcation υX is a Lindelöf Σ-space. We are ready to prove the following theorem [238].

Theorem 9.15 (A) For a topological space X, the following assertions are equivalent: (i) υX is a Lindelöf Σ-space. (ii) X is strongly web-bounding. (iii) Cp(X) is web-bounded. (iv) Lp(X) is web-bounded. (v) Cp(X) is a dense subspace of an lcs that is a Lindelöf Σ-space. (B) For a topological space X, the following conditions are equivalent: (i) Cp(X) is web-bounded and X is realcompact. (ii) Lp(X) is a Lindelöf Σ-space.

Proof (A) (ii) ⇒ (i): Assume ﬁrst that X is strongly web-bounding, and let {Aα : α ∈ Σ} be a covering of X verifying the web-bounding condition. Then, for each ∈ = ∈ ∈ N ⊂ f C(X) and each α (nk)k∈N Σ, there exists k such that f(Cn1n2···nk ) [−k,k].Theset

:= { ∈ RX :∀ = ∈ ∃ ∈ N ⊂[− ]} Z f α (ni) Σ, k ,f(Cn1n2···nk ) k,k verifies the condition X Cp(X) ⊂ Z ⊂ R . (9.8) Endow Z with the topology induced by RX.LetR = R ∪{−∞, +∞} be the two- X point compactification of R homeomorphic to [0, 1]. Then R is a compactification = ∈ ∈ N = of Z. For each α (ni)i∈N Σ and k ,letFα|k Fn1n2···nk be the closure in X R of the set { ∈ RX : ⊂[− ]} f f(Cn1n2···nk ) k,k . The family

~~S := {Fα|k,α∈ Σ,k ∈ N} X is a countable family of compact subsets of R . Clearly, X X R \Z = R \RX ∪ RX\Z . 9.3 Bounded resolutions for Cp(X) 223~~

~~Take arbitrary~~

RX g ∈ Z \Z. X If g ∈ R \RX, there exists a ∈ X such that g(a) ∈{−∞, +∞}. There exists α = (ni) ∈ Σ such that a ∈ Aα. Then, from ∩{−∞ +∞} = ∅ g(Cn1n2···nk ) , , it follows that g/∈ Fα|k for each k ∈ N. X If g ∈ R \Z, there exists α = (ni) ∈ Σ such that for each k ∈ N we have [− ] g(Cn1n2···nk ) k,k .

X Also,wehavethatg/∈ Fα|k for each k ∈ N. Therefore, if f ∈ Z and g ∈ R \Z, there exists α = (ni) ∈ Σ such that g/∈ Fα|k for each k ∈ N. From the deﬁnition of Z, it follows that for this α there exists k ∈ N such that f ∈ Fα|k. Therefore, by Proposition 9.16, we deduce that Z is a Lindelöf Σ-space. Finally, we apply Proposition 9.5 to get that υX is a Lindelöf Σ-space. (i) ⇒ (ii): Assume υX is a Lindelöf Σ-space. Then there exists Σ ⊂ NN and an usco compact-valued map T from Σ into υX covering υX. Set Aα := T(α)∩ X = ∈ ∈ for α (nk) Σ. Take a sequence fk Cn1,n2,...,nk . There exists a sequence (αk)k in Σ that converges to α such that fk ∈ T(αk) for each k ∈ N. Since T is usco, the set {fk : k ∈ N} is countably compact and hence topologically bounded. (iii) ⇔ (i): To prove this equivalence, in (9.8) replace X by Cp(X).IfCp(X) is strongly web-bounding, then

Cp(X) Cp(Cp(X)) ⊂ Z ⊂ R and Z is Lindelöf Σ. If we need to assume that Cp(X) is only web-bounded, one should have more space for Z to be a Lindelöf Σ-space. Indeed, if Cp(X) is web- bounded, we deduce (analogously) that there exists a Lindelöf Σ-space Z such that

~~Cp(X) Lp(X) ⊂ Z ⊂ R .~~

C (X) Since X ⊂ Lp(X), then X ⊂ Z ⊂ R p . Now the classical procedure (see the proof of [27, Theorem IV.9.5]) applies to show that υX is a Lindelöf Σ-space. Indeed, if Y is the closure of X in Z, the space Y is a Lindelöf Σ-space. Since every real-valued function on X can be continuously extended to RCp(X), then υX = υY = Y is a Lindelöf Σ-space. Conversely, if υX is a Lindelöf Σ-space, by Proposition 9.5 there exists a Lindelöf Σ-space Z such that

~~X Cp(X) ⊂ Z ⊂ R .~~

~~Then clearly Cp(X) is web-bounded. To prove (iii) ⇔ (v) in (A), it is enough to apply Claim 9.1 below. 224 9 K-analytic and Analytic Spaces Cp(X)~~

Claim 9.1 For an lcs E, the following conditions are equivalent: (a) E is web-bounded. (b) The space (E, σ (E, E)) is embedded in a locally convex Lindelöf Σ-space (W, σ (W, E)), where E ⊂ W ⊂ (E)∗. (c) (E,σ(E,E))is web-bounded. (d) The space (E,σ(E,E))is embedded in a locally convex Lindelöf Σ-space (Z, σ (Z, E)), where E ⊂ Z ⊂ E∗.

Indeed, (a) ⇒ (d): Assume E is web-bounded and that {Aα : α ∈ Σ} is a = ∈ ∈ covering of E such that, if α (nk) Σ and xk Cn1n2···nk , then (xk)k is bounded. Clearly, for each α ∈ Σ and each x ∈ E, there exists k ∈ N such that ⊂[− ] x (Cn1n2···nk ) k,k . Set := { ∈ ∗ :∀ = ∈ ∃ ∈ N ⊂[− ]} Z x E α (ni) Σ, k ,x (Cn1n2···nk ) k,k . ∗ Since (Cn1,n2,...,nk )k is decreasing, Z is a vector subspace of E and

~~E ⊂ Z ⊂ E∗ ⊂ RE.~~

Similarly, as we proved in (A), we deduce that (Z, σ (Z, E)) is a Lindelöf Σ-space. (d) ⇒ (c) is obvious. (c) ⇒ (b): By the hypothesis, (E,σ(E,E)) is web-bounded, and if we apply to this space the implication (a) ⇒ (d), we note that the weak dual (E, σ (E, E)) is embedded in a locally convex Lindelöf Σ-space (W, σ (W, E)). (b) ⇒ (a) is trivial. Claim 9.1 is proved. We continue the proof of Theorem 9.15. (iii) ⇔ (iv): Since Lp(X) = Cp(X), we apply Claim 9.1 to prove the equivalence. (B) Assume Cp(X) is web-bounded with realcompact X. By (A), the space X = υX is a Lindelöf Σ-space. Then, by Proposition 6.13, the space Lp(X) is a Lindelöf Σ-space. This proves (i) ⇒ (ii). For the converse, if Lp(X) is a Lindelöf Σ-space, the space X ⊂ Lp(X) (as a closed subspace) is a Lindelöf Σ-space. Finally, by (A), the space Cp(X) is web- bounded.

~~We will need also the following general fact.~~

~~Lemma 9.6 If E is an lcs such that (E, σ (E, E)) has countable tightness, the weak∗ dual (E,σ(E,E))is a realcompact space.~~

Proof By Proposition 12.1 below, it is enough to show that every linear functional f on E that is σ(E,E)-continuous on each σ(E,E)-closed separable vector subspace is continuous on E. We prove that the kernel K := {x ∈ E : f(x)= 0} of f is a closed subspace in E, and this yields the continuity of f . Indeed, if y ∈ K, there exists a countable set D ⊂ K such that y ∈ D (where the closure is taken in σ(E,E)). 9.3 Bounded resolutions for Cp(X) 225

~~By the assumption, the restriction map f |span(D) is σ(E,E)-continuous. This shows that f(y)∈ f(span(D)) ⊂ f(K)={0}. Hence y ∈ K and f ∈ E.~~

~~Note the following proposition; see also [401, Proposition 2.11].~~

~~Proposition 9.17 Let X be a Lindelöf P-space. Then X is countable if and only if Lp(X) has countable tightness if and only if Cp(X) is realcompact.~~

Proof First assume that X is countable. Then Cp(X) is a separable metrizable lcs. Hence Lp(X) ⊂ Cp(Cp(X)) has countable tightness, again by Proposition 9.9. ∗ Now assume that the weak dual Lp(X) of Cp(X) has countable tightness. By Lemma 9.6, the space Cp(X) is a realcompact space. Since X is a P-space, the topological group Cp(X, [−1, 1]) is a countably compact space; see, for example, [31]. Consequently, its closure K in (Cp(X)) is compact (as Cp(X) is realcompact). Moreover, K has countable tightness by Proposition 9.9. Since Cp(X) embeds in Cp(X, [−1, 1]) (take, for example, f → f/(1 +|f |), f ∈ C(X)), we apply Propo- sition 9.11 to deduce that Cp(X) is metrizable. Hence X is countable.

~~Since Cp(X) is angelic if υX is a Lindelöf Σ-space (Theorem 4.5 and Lemma 9.2), Theorem 9.15 yields the following corollary.~~

~~Corollary 9.9 Every web-bounded space Cp(X) is angelic.~~

Since every topological space with a compact resolution is quasi-Suslin and X is closed in Lp(X), an immediate consequence of Theorem 9.15, Proposition 6.13 combined with Proposition 9.15, implies the following dual version of Theorem 9.9.

~~Corollary 9.10 The space Lp(υX) is almost quasi-Suslin if and only if it is a Lindelöf Σ-space. The space Lp(υX) is K-analytic if and only if it has a compact resolution.~~

~~Now we prove the following theorem.~~

~~Theorem 9.16 For a Baire lcs E, the following statements are equivalent: (i) E is metrizable. (ii) (E,σ(E,E))is an almost quasi-Suslin space. (iii) E is web-bounded.~~

Proof (i) ⇒ (ii): If (Un)n is a decreasing basis of absolutely convex neighborhoods := ◦ of zero inE, its polars Kn (Un) are σ(E ,E)-compact sets in (E ,σ(E ,E)) = and E n Kn. Therefore (E ,σ(E ,E))is σ -compact. 226 9 K-analytic and Analytic Spaces Cp(X)

⇒ (ii) (iii): Since the realcompactiﬁcation υ(Eσ ) is a Lindelöf Σ-space, by := Theorem 9.15 the space Cp(Eσ ) is web-bounded, where Eσ (E ,σ(E ,E)).By ⊂ (E, σ (E, E )) Cp(Eσ ), the space E is a web-bounded lcs. N (iii) ⇒ (i): By the assumption, there exists a set Σ ⊂ N and a family {Aα : α ∈ Σ} covering E such that for each α = (nk) ∈ Σ and each

~~∈ xk Cn1,n2,...,nk the sequence (xk)k is bounded in E. Therefore, for each neighborhood of zero W and each α = (nk) ∈ Σ, there exists r ∈ N such that Cr ⊂ rW, where Cr :=~~

Cn1,n2,...nr . Since = = E Cn1 ,Cn1 Cn1,n2 ,..., n1 n2 and since E is a Baire space, there exists α = (nk) ∈ Σ such that Ck − Ck is a neighborhood of zero for each k ∈ N.LetU and V be closed, absolutely convex neighborhoods of zero such that V − V ⊂ U. There exists k ∈ N such that

~~Ck − Ck ⊂ kV − kV ⊂ kU.~~

~~Hence the sets −1 Wk := k (Ck − Ck) compose a countable basis of neighborhoods of zero in E,soE is metrizable.~~

~~:= Corollary 9.11 Let (Ei)i∈I be a family of nonzero web-bounded lcs’s. Then E | |≤ℵ i∈I Ei is web-bounded if and only if I 0.~~

Proof Assume I is uncountable and E is web-bounded. Then E contains a subspace of the form RA for some uncountable set A. Endow A with the discrete topology. A A A Since R = Cp(A), the Baire space R is web-bounded. By Theorem 9.16, R is metrizable, so A is countable, a contradiction.

= Corollary 9.12 Let E t∈T Et be the product of Fréchet spaces. Then T is countable if and only if the space (E,σ(E,E)) is K-analytic if and only if υ(E,σ(E,E))is a Lindelöf Σ-space if and only if (E,σ(E,E))is almost quasi- Suslin.

Proof Assume T is uncountable and Eσ is almost quasi-Suslin. Then υEσ is a Lindelöf Σ-space by Proposition 9.15. By Claim 9.1 in Theorem 9.15, we deduce that E is web-bounded, and then we apply Theorem 9.16 to reach a contradiction.

In order to prove Proposition 9.20, we need the following lemma due to Tkachuk [400]. 9.3 Bounded resolutions for Cp(X) 227 = ∈ Lemma 9.7 Assume that Cp(X) n Cn. Then there exists f Cp(X), ε>0 and n ∈ N such that (Cn + f)∩ C(X,(−ε, ε)) is dense in Cu(X, (−ε, ε)) endowed with the uniform topology.

Proof For each n ∈ N,set := ∩ b Bn Cn Cp(X), b where Cp(X) is a subspace of Cp(X) of continuous and bounded functions on X. b = b Since Cu(X) n Bn and Cu(X) is a Fréchet space, by the Baire category theorem b there exist n ∈ N, g ∈ C (X) and ε>0 such that Bn is dense in

~~b Kε(g) := h ∈ C (X) : sup |h(x) − g(x)| <ε . x∈X~~

~~Then Bn + f is dense in Cu(X, (−ε, ε)), where f := −g. Consequently,~~

~~(Bn + f)∩ C(X,(−ε, ε)) ⊂ (Cn + f)∩ C(X,(−ε, ε)) is dense in Cu(X, (−ε, ε)). This completes the proof.~~

~~Lemma 9.7 yields the following two interesting corollaries [400].~~

~~Corollary 9.13 If Cp(X) is covered by a sequence (Cn)n of closed subsets, there exists m ∈ N such that Cm contains a subset homeomorphic to Cp(X).~~

~~Corollary 9.14 If Cp(X) is covered by a sequence of metrizable closed subsets, the space Cp(X) is metrizable.~~

Motivated by Theorem 9.12, one may ask about conditions for Cp(X) to have a fundamental sequence of bounded sets. Since Cp(X) is always a quasibarrelled space [213], this question is in fact to determine when Cp(X) is a (DF )-space. Corollary 9.13 is applied to deduce the following proposition.

~~Proposition 9.18 If Cp(X) admits a fundamental sequence of bounded sets, the space X is ﬁnite.~~

A family U is called an ω-cover of X if for any ﬁnite subset A of X there exists U ∈ U such that A ⊂ U. We will also use the following characterization of countable tightness of Cp(X);see[179], [284].

~~Proposition 9.19 Cp(X) has countable tightness if and only if every open ω-cover of X contains a countable ω-subcover.~~

~~If a topological space X has the property in Proposition 9.19, we say that X is an ω-space. The following result is due to Tkachuk [400]. 228 9 K-analytic and Analytic Spaces Cp(X)~~

~~Proposition 9.20 (Tkachuk) If Cp(X) is covered by a sequence (Cn)n of subsets of countable tightness, then the space Cp(X) has countable tightness.~~

~~Proof By Lemma 9.7, there exist n ∈ N, ε>0, and f ∈ Cp(X) such that~~

~~D := (f + Cn) ∩ C(X,(−ε, ε)) is dense in Cu(X, (−ε, ε)). We show that every open ω-cover in X contains a countable ω-subcover. Let U be an ω-cover of nonempty open sets in X. Set~~

~~− − P := {f ∈ D :∃U ∈ U ,f 1(10 1ε, ε) ⊂ U}.~~

Note that D ⊂ P , the closure in Cp(X). Indeed, choose arbitrary l ∈ D, δ>0, and a ﬁnite set x1,x2,...,xn in X.Theset n V(l):= {f ∈ Cp(X) :|l(xi) − f(xi)| <δ} i=1 is a neighborhood of l in Cp(X). We need to show that there exists h ∈ P contained in V(l). There exists U ∈ U such that

~~{x1,x2,...,xn}⊂U.~~

For each 1 ≤ i ≤ n,setri := l(xi). There exists a function g ∈ Cp(X, (−ε, ε)) such that g(xi) = ri for each 1 ≤ i ≤ n and g(X \ U) ={0}. Since D is dense in Cu(X, (−ε, ε)), there exists h ∈ D such that − sup |h(x) − g(x)| < min{δ,10 1ε}. x∈X

Then clearly h ∈ P . On the other hand, h ∈ V(l). The claim is proved. −1 Next, choose a function f0 ∈ D such that f0(x) > (10) 9ε for each x ∈ X. Then f0 ∈ P . By the assumption, there exists a countable subset K ⊂ P such that f0 ∈ K. Since K ⊂ P , for each s ∈ K we choose a set Us ∈ U such that

~~−1 s(10 ε, ε) ⊂ Us.~~

Set V := {Us : s ∈ K}. Clearly, the family V is countable. The proof will be completed if we show that V is a ω-cover. Let A := {y1,y2,...,yn}⊂X. There exists f1 ⊂ K such that −1 |f1(yi) − f0(yi)| < 10 ε for each 1 ≤ i ≤ n. Then

~~−1 −1 f1(yi)>(10) 8ε>(10) ε ≤ ≤ ⊂ V for each 1 i n. This means that A Uf1 . We proved that is a ω-cover. 9.3 Bounded resolutions for Cp(X) 229~~

~~From Theorem 9.15 (B) and Proposition 9.9, we know that, if Cp(X) is web- bounded with countable tightness, Lp(X) is a Lindelöf Σ-space. On the other hand, we have the following example.~~

~~Example 9.4 Assume the CH. There exists a compact space X of cardinality ℵ1 such that Cp(X) is Lindelöf and not a Lindelöf Σ-space, and Lp(X) is web- bounded with countable tightness.~~

Proof Under the CH, Kunen constructed a compact scattered, hereditarily separable space K of the cardinality ℵ1 such that the Banach space Cc(K) is weakly (hereditarily) Lindelöf; see [347], [336]. Since K is a zero-dimensional space, by n [27, Theorem IV.8.6] any ﬁnite product Cp(K) is a Lindelöf space. Therefore, by Proposition 9.9, the space Lp(K) ⊂ Cp(Cp(K)) has countable tightness. We prove that υCp(K) = Cp(K) is not a Lindelöf Σ-space. As K is separable, it is enough to prove that it does not have a countable network (or equivalently is not cosmic) because of Proposition 9.3. Assume K is a cosmic space. Then Cp(K) also is cosmic by Proposition 9.2. Hence Cp(K) is a separable space. Hence K admits a weaker metric topology. On the other hand, it is known that a metric compact scattered space is countable, a contradiction.

Recall again that a compact space X is scattered if every closed subset L ⊂ X has an isolated point in L. A point p is isolated in X if there exists a neighborhood U of p in X such that U ∩ X ={p}. We refer the reader to [268], where examples of Corson compact spaces X such that Cp(X) is not a Lindelöf Σ-space are presented; see also [4]. Example 9.5 (see [24]) provides additional restrictions on possible extensions of Theorem 9.15 (B).

~~Example 9.5 There exists a space Z that is not Lindelöf such that Cp(Z) is K- analytic, Lp(Z) has countable tightness and Lp(Z) is neither separable nor Lin- delöf.~~

Proof Let Y be a Talagrand compact space such that y ∈ Y , βZ = Y , where Z = Y \{y} and Z is pseudocompact and not compact; see [24, Example 7.14] or [149, Part 8.4]. Hence Cp(Z) is K-analytic and Z is not Lindelöf. Since Z is not Lin- delöf, the space Cp(Z) does not have countable tightness by Proposition 9.9.Also, the space Lp(Z) is not separable; otherwise there exists on Cp(Z) a weaker metric topology, and then Cp(Z), being K-analytic, must be analytic. Then, by The- orem 9.7, Z is σ -compact and hence Lindelöf, a contradiction. Clearly, Lp(Z) is not Lindelöf since it contains the closed subspace Z. On the other hand, Lp(Z) has countable tightness, since Lp(Z) ⊂ Cp(Cp(Z)), and Cp(Cp(Z)) has countable tightness. 230 9 K-analytic and Analytic Spaces Cp(X)

~~9.4 Some examples of K-analytic spaces Cp(X) and Cp(X, E)~~

In this section, we provide more examples of K-analytic spaces Cp(X), Lp(X, E) and Cp(X, E). Most results in this section are due to Canela [78]. For two lcs’s X and E,byLp(X, E) we denote the subspace of Cp(X, E) of linear maps from X into E. The topology of Lp(X, E) is the topology of the pointwise convergence, also called a simple topology; this space is sometimes denoted by Ls(X, E).As ∗ usual, Lp(X) means the weak dual of Cp(X). We start with the following simple observation.

~~Proposition 9.21 Let X be a separable metrizable lcs. Let E be a separable Fréchet space. Then Lp(X, E) is analytic.~~

Proof First assume that E is a separable Banach space. Let (Vn)n be a countable basis of neighborhoods of zero in X. Set V(Vn,B):= {f ∈ L (X, E) : f(Vn) ⊂ B}, where B denotes the unit ball in E, and the sets V(Vn,B) are endowed with the topology from Lp(X, E).If{xn : n ∈ N} is a countable dense subset of X,themap N T : V(Vn,B)→ E , deﬁned by f → (f (xn)n), is a homeomorphism onto a closed N L = subspace of E . Therefore V(Vn,B)is analytic. Hence p(X, E) n V(Vn,B) is analytic. Now assume that E is a separable Fréchet space. Then, since E is topologically isomorphic to a closed subset of a countable product n En of separable Banach spaces, Lp(X, E) is topologically isomorphic to a closed subspace of the L L analytic space n p(X, En). Hence p(X, E) is analytic.

~~Similarly, we prove the following proposition.~~

~~Proposition 9.22 Let X be a separable normed space. Let E be a complete K-analytic (analytic) lcs with a fundamental sequence of bounded sets. Then Lp(X, E) is K-analytic (analytic).~~

2 2 Example 9.6 Let (I)σ and c0(I)σ be the Banach spaces (I) and c0(I) endowed with the weak topologies, respectively. If I is uncountable, then the space 2 Cp( (I)σ ,c0(I)σ ) is not K-analytic and c0(I)σ is K-analytic.

2 2 Proof Note that the space Lp( (I)σ ,c0(I)σ ) equals Lp( (I), c0(I)σ ). Since the 2 2 space Lp( (I)σ ,c0(I)σ ) is a closed subspace of Cp( (I)σ ,c0(I)σ ), it is enough 2 to show that Lp( (I), c0(I)σ ) is not K-analytic; see [389]. Clearly, c0(N × I) is isomorphic to c0(I). 2 I Let (ei)i∈I be a canonical basis in (I). We deﬁne a homeomorphism T from N 2 I into a closed subspace of Lp( (I), c0(I)σ ).Fors = (si) ∈ N , deﬁne Ts(ei) := fi, where 1 j = i, n ≤ s , f (n, j) = i i 0 otherwise.

2 Then the map T : s → Ts is a homeomorphism. Assume that Lp( (I)σ ,c0(I)σ ) is K-analytic. Then NI is also K-analytic. Since NI is not Lindelöf, we reach a 9.5 K-analytic spaces Cp(X) over a locally compact group X 231 contradiction. The space c0(I) is a WCG Banach space, and hence c0(I)σ is K- analytic; see Theorem 12.8 below.

~~Example 9.7 If X is Eberlein compact and E is a nuclear Fréchet space, Cp(X, Eσ ) is K-analytic.~~

Proof Since X is Eberlein compact, X is homeomorphic with a weakly compact subset of a Banach space. Since the weak topology of a metrizable lcs is angelic, X is angelic. Hence X is Fréchet–Urysohn. On the other hand, since E is a nuclear Fréchet space, E is a closed subspace of a countable product of the space 1 so E has the Schur property (i.e., every weakly convergent sequence in E converges in the original topology of E; see also [364, Corollary 2, p. 101]). As every sequentially continuous map on a Fréchet–Urysohn space is continuous, we deduce that 1 C(X,Eσ ) = C(X,E). The space Cp(X, ) is K-analytic by Example 11.1. Since 1 Cp(X, E) is closed in a countable product of the space Cp(X, ), and the topology of Cp(X, Eσ ) is equal to the original one of Cp(X, E), the conclusion follows.

~~9.5 K-analytic spaces Cp(X) over a locally compact group X~~

It is still unknown (see [24, Problem 44, p. 29]) when exactly for a given X the space Cp(X) is a Lindelöf space. It is known [146, 3.8.D] that for any second countable X the space Cp(X) is Lindelöf. The same holds for (not necessarily second countable) Corson compact spaces X [27, Theorem IV.2.22]. We refer the reader to [21], [24], [76], [91] and [135] for more known results concerning this problem. Theorem 9.17, published in [234], collects several equivalent conditions for Cp(X) to be a Lindelöf space for a locally compact group X.

Theorem 9.17 For a locally compact topological group X, the following assertions are equivalent: (1) Cp(X) is analytic. (2) Cp(X) is K-analytic. (3) Cp(X) is Lindelöf. (4) X is metrizable and σ -compact. (5) X is analytic. (6) Cp(X) has a bounded resolution and X is metrizable. (7) Cc(X) is a separable Fréchet space. (8) Cc(X) has a compact resolution. (9) Cc(X) has a bounded resolution and X is metrizable.

~~We need the following two propositions; for the ﬁrst one, see [108] and also [100, Theorem 1 and Remark (ii)]. 232 9 K-analytic and Analytic Spaces Cp(X)~~

Proposition 9.23 Let X be a locally compact topological group. Then there exist a compact subgroup G of X, n ∈ N ∪{0}, and a discrete subset D ⊂ X such that X is homeomorphic to the product Rn × D × G.

Proposition 9.24 For a locally compact topological group X, the following assertions are equivalent: (i) X is angelic. (ii) Every compact subgroup of X has countable tightness. (iii) X is metrizable. (iv) X has countable tightness.

Proof The only nontrivial implication is (ii) ⇒ (iii): Assume ﬁrst that X is compact. By Kuz’minov’s theorem [263, Theorem], every compact Hausdorff group X is dyadic (i.e., X is a continuous image of {0, 1}α, where α is some cardinal number). Since every dyadic Hausdorff space with countable tightness is metrizable ([146, 3.12.12(h)], [19, Theorem 3.1.1]), the conclusion holds. Now assume that X is a locally compact group. The previous case combined with Proposition 9.23 completes the proof.

~~Now we are ready to prove Theorem 9.17.~~

Proof Clearly, (1) ⇒ (2) ⇒ (3). (3) ⇒ (4): Assume that Cp(X) is Lindelöf. By Theorem 9.11, every ﬁnite product Xn of X has countable tightness. By Proposition 9.24, the space X is metrizable. This proves that X is a metrizable space and separable by Proposition 9.23. Indeed, D note that Cp(X) is continuously mapped onto Cp(D),soCp(D) = R is Lindelöf. Hence D is countable. (4) ⇒ (5)isclear. (5) ⇒ (4): Since X is an analytic Baire topological group, Theorem 7.3 is applied to get that X is metrizable and separable. (4) ⇒ (6): Since X is locally compact and σ -compact, it is hemicompact, and then Cc(X) is a metrizable lcs. Hence Cc(X) has a bounded resolution, and consequently Cp(X) also has a bounded resolution. (6) ⇒ (4): Since X is a metrizable locally compact topological group, X is paracompact, and it is σ -compact by Corollary 7.2. (4) ⇒ (7): If X is locally compact, metrizable and σ -compact, the space Cc(X) is a separable Fréchet space. (7) ⇒ (8)isobvious. (8) ⇒ (9): Since Cc(X) is trans-separable by Proposition 6.10, every compact subset of X is metrizable; see again Proposition 6.5. Now Proposition 9.23 is applied to get that X is metrizable. (7) ⇒ (1)isclear. We need to prove (9) ⇒ (4): Since X is paracompact and locally compact, the space Cc(X) is Baire by Proposition 2.2. Now Proposition 7.1 shows that Cc(X) is metrizable. Hence X is hemicompact, so (4) holds, too, and the proof is complete. 9.5 K-analytic spaces Cp(X) over a locally compact group X 233

We note a few remarks related to Theorem 9.17. (1) Arkhangel’skii asked if every compact homogeneous Hausdorff topological space with countable tightness is first-countable. Dow [136, Theorem 6.3] answered this question positively under the Proper Forcing Axiom. Although Eberlein compact spaces provide a large class of spaces with countable tightness and it is known that homogeneous Eberlein compact spaces are first-countable [27, III.3.10], nonmetrizable homogeneous Eberlein compact spaces exist [296]. On the other hand, Gruenhage showed [192] that every Gul’ko compact space contains a dense Gδ- subset that is metrizable and hence each compact group that is Gul’ko compact is metrizable. Nevertheless, there are Corson compact spaces without any dense metrizable subspace [407]. Since every Corson compact space has countable tightness [219], Proposition 9.24 yields that in the class of topological groups the Eber- lein, Talagrand, Gul’ko and Corson compactnesses are equivalent, and each such compact group is metrizable. (2) A locally compact topological group X is metrizable and σ -compact if and only if Cc(X) is weakly Lindelöf. (3) From Theorem 9.17, it follows that the analyticity, K-analyticity and Lindelöf property for Cc(X) over a locally compact topological group X are equivalent conditions. Moreover, for a locally compact topological group X, the space Cp(X) is analytic (K-analytic) if and only if Cc(X) is weakly analytic (weakly K-analytic). (6) Note that ((i)+(ii)+(iv)) ⇒ (iii) in Proposition 9.24 fails for topological groups that are not locally compact. Indeed, let X be an infinite-dimensional reflexive separable (real) Banach space endowed with the weak topology. Then X is a σ -compact angelic lcs (Corollary 4.3) whose compact subsets are metrizable (Corollary 4.6) has countable tightness (Theorem 12.3 below) and is not metrizable. (7) Note also that the group structure in (iv) ⇒ (i) in Proposition 9.24 is essential. The one-point compactification of the space Ψ of Isbell is an example of a compact Hausdorff space with countable tightness that is not angelic; see [165, pp. 54–55]. It is known that there are nonmetrizable absolutely convex weakly compact sets in Banach spaces over the field of real or complex numbers. This situation is different if the valued field K is non-Archimedean and locally compact. Recall that a nontrivially valued field K := (K, |·|) is non-Archimedean if |t + s|≤max {|t|, |s|} for all t,s ∈ K;see[348]. A subset B of a vector space E over a non-Archimedean nontrivially valued field K is called absolutely K-convex if from x,y ∈ B, t,s ∈ K, and |t|≤1, |s|≤1 it follows that tx+ sy ∈f B.I E is a tvs over a non-Archimedean nontrivially valued complete field K and E contains a nonzero compact absolutely K-convex set, K must be locally compact; see [126]. We note the following proposition.

Proposition 9.25 Let E be a tvs over a locally compact nontrivially valued ﬁeld K. Then: (i) If K is Archimedean, every locally compact subgroup X of E is metrizable. (ii) If K is non-Archimedean and E is metrizable, every absolutely K-convex locally compact subset X of E,σ(E,E) is metrizable in σ(E,E). 234 9 K-analytic and Analytic Spaces Cp(X)

Proof (i) Since K is Archimedean, by Ostrowski’s theorem [348, Theorem 1.2] K is the ﬁeld of either the real or complex numbers. By the assumption, X is locally compact, so it is homeomorphic to the product Rn × D × G, where D and G are as in Proposition 9.23. As any compact subgroup in a (real or complex) topological vector space is trivial, the conclusion follows. (ii) Since (K, +) is a locally compact Abelian group, K∧ separates points of K. Fix a nonconstant χ ∈ K∧. Then E∧ ={χ ◦ x : x ∈ E} by [414, Theorem 2]. As E =: Hom(E, T), where T denotes the torus in the complex plane, separates points of E (see [366]), we deduce that the Bohr topology of the group (E, +) is Haus- dorff. The equality E∧ ={exp(if ) : f ∈ E} ensures that σ(E,E∧) ≤ σ(E,E). E is metrizable, so by Proposition 4.5 we deduce that (E, σ (E, E∧)) is angelic. Next, we apply Corollary 4.2 to see that (E, σ (E, E)) is angelic. On the other hand, since K is non-Archimedean and X is an absolutely K-convex subset of E, X is an additive subgroup of E. Hence X is an angelic locally compact group, and by Proposition 9.24 we note that it is metrizable.

~~Example 9.8 Proposition 9.25 (i) fails for a non-Archimedean K.~~

c Proof Set K := Q2 and B := {α ∈ Q2 :|α|2 ≤ 1}. Then B is a nonmetrizable compact additive subgroup of the topological vector space Kc. Using the same example, we deduce that the metrizability assumption of E is essential in Proposition 9.25 (ii).

~~Since the weak topology σ(E,E) of a metrizable lcs E is angelic and has countable tightness, Proposition 9.25 may suggest the following problem.~~

~~Problem 9.2 Let E be a real lcs. For which compact subsets X of E does there exist a compact topological group that is homeomorphic to X?~~

~~Note that every such X must be homogeneous. Also, X cannot be convex since the Schauder ﬁxed-point theorem fails for compact topological groups (translations do not have a ﬁxed point).~~

∧ 9.6 K-analytic group Xp of homomorphisms This section deals with a variant of Theorem 9.17 for the group of homomorphisms ∧ Xc . For Abelian topological groups X and Y ,byHomp(X, Y ) and Homc(X, Y ) we denote the set Hom(X, Y ) of all continuous homomorphisms from X into Y endowed with the pointwise and compact-open topologies, respectively. Set ∧ =: T ∧ =: T Xp Homp(X, ), Xc Homc(X, ), where T denotes the unit circle of the complex plane. For every x ∈ X, the function x∧ : X∧ → T, deﬁned by x∧(f ) := f(x) for f ∈ X∧, is a continuous homo- ∧ { ∧ : ∈ }⊂ ∧ ∧ morphism on Xc , and x x X (Xc ) , and by the Pontryagin–van Kampen ∧ 9.6 K-analytic group Xp of homomorphisms 235 theorem (see [205, Theorem 24.8]), the map x → x∧ is a topological isomorphism ∧ ∧ between a locally compact Abelian group X and (Xc )c .IfX is an Abelian locally ∧ compact group, Xc is also locally compact and Abelian, and by the Peter–Weyl– ∧ van Kampen theorem [304, Theorem 21] the space Xc is dual-separating (i.e., for different x,y ∈ X, there exists f ∈ X∧ such that f(x)= f(y)). For an Abelian group X, the set of all homomorphisms from X into T endowed with the pointwise convergence topology is a compact Abelian group, as a closed subgroup of the product TX;see[206, Proposition 1.16]. For a metrizable Abelian ∧ topological group X, the group Xc is always an Abelian, Hausdorff complete and hemicompact group. Moreover, it is a k-space; see [37, Corollary 4.7] and [93].

~~Proposition 9.26 If X is a separable and metrizable Abelian topological group, ∧ ∧ Xc is locally compact if and only if Xc is metrizable.~~

∧ : × ∧ → T Proof If Xc is metrizable, the evaluation map e X Xc is continuous. Then, ∧ by [281, Proposition 1.2], the group Xc is locally compact. To prove the converse, ∧ note that every compact subset of Xc is metrizable (since X is separable), and Proposition 9.24 applies.

~~We also need a couple of additional results.~~

~~Proposition 9.27 A locally compact Lindelöf topological group X is hemicompact.~~

ProofTake an open neighborhood U of the unit whose closure U is compact. Since = X x∈X xU and X is a Lindelöf space, there exists a sequence (xn)n such that = := n X n xnU.Set Kn i=1 xiU.Then (Kn)n is a fundamental sequence of compact sets in X.

~~∧ Proposition 9.28 If X is a metrizable, locally compact Abelian group, Xc is a hemicompact k-space.~~

Proof Let (Un)n be a decreasing basis of neighborhoods of the unit in X. Then := { ∈ ∧ : ⊂ T } Un φ Xc φ(Un) + is compact in the compact-open topology for each n ∈ N, where T+ := {z ∈ T : ≥ } ∧ = ∧ Re z 0 . Moreover, X n Un . If K is a compact set in the space Xc , ⊂ ∧ ∧ K (Xc ) is a neighborhood of the unit. By the Pontryagin–van Kampen theorem [205], it can be identiﬁed with

~~K := {x ∈ X : Re φ(x) 0,φ∈ K}.~~

~~Since the last set is a neighborhood of the unit, Um ⊂ K for some m ∈ N. Hence ⊂ ⊂ K K Um . 236 9 K-analytic and Analytic Spaces Cp(X)~~

~~∧ For the proof that Xc is a k-space, we refer to [37, Corollary 4.7].~~

~~[37, Proposition 2.8] implies the following proposition.~~

~~∧ Proposition 9.29 If a topological Abelian group X is hemicompact, Xc is metrizable.~~

~~We also need the following simple proposition.~~

~~Proposition 9.30 If Cp(X, R) has countable tightness, Cp(X, Y ) has countable tightness for any metric space (Y, d).~~

Proof Let A ⊂ Cp(X, Y ). Assume f ∈ A. Deﬁne a continuous map T : Cp(X, Y ) → Cp(X, R) by T (g)(x) := d(g(x), f (x)), where g ∈ Cp(X, Y ), x ∈ X. Note that 0 = T(f)∈ T(A) ⊂ T(A).By the assumption, there exists in A a countable subset B such that T(f)∈ T(B); hence f ∈ B.

~~Now we are ready to prove the following main result of [237].~~

Theorem 9.18 Let X be a locally compact Abelian group. The following assertions are equivalent: (1) X is metrizable. ∧ (2) Xp is σ -compact. ∧ (3) Xp is K-analytic. (4) (X, σ (X, X∧)) has countable tightness. Moreover, if X is Lindelöf, any condition above is equivalent to ∧ (5) Xc is metric, complete and separable. ⇒ ∧ ∧ Proof (1) (2): By Proposition 9.28, the group Xc is hemicompact, so Xp is σ -compact. ⇒ ∧ (2) (3): For an increasing sequence (Bn)n of compact sets covering Xp ,set N T(α):= Bn for α = (nk) ∈ N . Clearly, T is an usco compact-valued map with 1 ∧ values covering Xp . ⇒ ∧ ∧ n (3) (4): Since Xp is K-analytic, any ﬁnite product (Xp ) is Lindelöf. ∧ R By Proposition 9.9, the space Cp(Xp , ) has countable tightness. Now Propo- ∧ C sition 9.30 is applied to say that Cp(Xp , ) has countable tightness. Hence ∧ ∧ C (X, σ (X, X )) (as topologically included in Cp(Xp , )) has countable tightness. (4) ⇒ (1): Since X is a locally compact group, there exist a compact subgroup G of X, n ∈ N ∪{0}, and a discrete subset D ⊂ X such that X is homeomorphic to the product Rn × D × G (see Proposition 9.23). Therefore the induced topology σ(X,X∧)|G coincides with the original one of G. Hence G has countable tightness. A compact group with countable tightness is metrizable by Proposition 9.24; hence X is metrizable. The remaining part follows from Proposition 9.27, Proposition 9.28 and Propo- sition 9.29. ∧ 9.6 K-analytic group Xp of homomorphisms 237

A topological space is sequential if every sequentially closed subset of X is ∧ closed. For a metrizable topological group X, the dual group Xc is Fréchet– ∧ Urysohn if and only if Xc is locally compact and metrizable; see, for example, [100, Theorem 2.1]. This provides a large class of complete angelic and hemicompact sequential groups that are not Fréchet–Urysohn [100, Theorem 2.3]. Glicksberg’s theorem states that for a locally compact topological group X the compact sets in X and (X, σ (X, X∧)) coincide; see [37].

~~Corollary 9.15 If X is a metrizable, locally compact, noncompact Abelian group, the group (X, σ (X, X∧)) has countable tightness and cannot be sequential, so neither can it be Fréchet–Urysohn.~~

Proof By Glicksberg’s theorem, the space (X, σ (X, X∧)) has the same compact subsets as X. Since X is metrizable, X is a k-space and X does not admit another k-space topology with the same compact sets. Therefore (X, σ (X, X∧)) is not a k-space and in particular cannot be sequential or Fréchet–Urysohn. Chapter 10 Precompact Sets in (LM)-Spaces and Dual Metric Spaces

Abstract This chapter presents uniﬁed and direct proofs of Pﬁster, Cascales and Orihuela and Valdivia’s theorems about metrizability of precompact sets in (LF )- spaces, (DF )-spaces and dual metric spaces, respectively. The proofs do not require the typical machinery of quasi-Suslin spaces, upper semicontinuous compact-valued maps and so on.

~~10.1 The case of (LM)-spaces: elementary approach~~

Floret [166] (motivated by results of Grothendieck, Fremlin, De Wilde and Pryce) proved an extended version of the Eberlein–Šmulian theorem with many applications. Nevertheless, Floret’s result said nothing about the metrizability of compact sets. Cascales and Orihuela [81] (answering a question of Floret [165]) showed that the weight of any precompact set in an (LM)-space is countable. Pﬁster and Val- divia proved the same result for (DF )-spaces and dual metric spaces, respectively [329], [423,Note4].K˛akol and Saxon [229] presented alternative proofs for (LM)- spaces and dual metric spaces. In this chapter, we present elementary proofs (mostly due to Kakol ˛ and Saxon [229]) of results mentioned above that do not require typical machinery involving usco maps and so on, showing that precompact sets in (LM)-spaces, dual metric spaces and (DF )-spaces are metrizable. If A ⊂ E is a subset of an lcs E,byac(A) and ac(A) we mean the absolutely convex and the closed absolutely convex envelopes of the set A, respectively. Recall again, for convenience, that (DF )-spaces are ℵ0-quasibarrelled spaces having a fundamental sequence of bounded sets (see the text below Proposi- tion 2.25). An lcs E is ∞-quasibarrelled if every β(E,E)-bounded sequence in E is equicontinuous. An lcs E is said to be dual metric if E has a fundamental sequence of bounded sets and every β(E,E)-bounded sequence in E is equicontinuous. Ev- ery semi-Montel dual metric space is a Montel (DF )-space [229]. An lcs E is a Montel (DF )-space if and only if E is a Montel (LB)-space if and only if E is the strong dual of a Fréchet–Montel space [229]. We need the following lemma.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 239 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_10, © Springer Science+Business Media, LLC 2011 240 10 Precompact Sets in (LM)-Spaces and Dual Metric Spaces

Lemma 10.1 Let P be a precompact set for the inductive limit space E with a deﬁning sequence (En)n of lcs. Assume that Un is an absolutely convex neighborhood of zero in En with Un ⊂ Un+1 for each n ∈ N. Then there exist m ∈ N and a ﬁnite subset F ⊂ P such that P ⊂ F + Um, where the closure is taken in E.

~~Proof Assume the conclusion fails. Then there exists a sequence (xn)n in P such that~~

xn+1 ∈{/ x1,x2,...,xn}+Un for each n ∈ N. Since the right-hand set above is closed in E, for each n ∈ N there exists a decreasing sequence (Vn)n of closed, absolutely convex neighborhoods of zero Vn in E such that

~~xn+1 ∈{/ x1,x2,...,xn}+Un + Vn.~~

~~Then Uk ∩ Vk ⊂ Un + Vn for all k,n ∈ N. This implies that V := ac Uk ∩ Vk ⊂ Un + Vn k for all n ∈ N. Hence~~

xn+1 ∈{/ x1,x2,...,xn}+V for all n ∈ N. Clearly, V is a neighborhood of zero in E. We proved that, for each ﬁnite subset F of {xn : n ∈ N},theset{xn : n ∈ N} is not included in F +V , showing that P is not precompact, a contradiction.

~~Lemma 10.1 is applied to get the following theorem.~~

Theorem 10.1 Let P be a precompact set for the inductive limit space E with a deﬁning sequence (En)n of lcs. Let Wn be a basis of absolutely convex neighborhoods of zero in En for each n ∈ N. Then the induced topology from E onto P has W |W |≤ |W | a basis of neighborhoods such that supn n .

~~Proof Assume that each Wn is inﬁnite. For each n ∈ N, select Wn ∈ Wn and set u := (Wn)n.LetU be the set of all such u.Foru ∈ U ,set −1 Un := 3 ac Wi . i≤n~~

~~By Lemma 10.1, there exist a ﬁnite subset Fu ⊂ P and nu ∈ N such that P ⊂ Fu + −1 := 3 Tu, where Tu Unu . Set~~

~~W := {P ∩ (x + Tu) : u ∈ U ,x∈ Fu}. W Since each set Fu is ﬁnite and Tu is described by a ﬁnite subset of the union n n, we have~~

~~|W |≤sup |Wn|. n 10.2 The case of dual metric spaces: elementary approach 241~~

We need to show that W is a basis of neighborhoods for P .Fixy ∈ P .LetU be a closed, absolutely convex neighborhood of zero in E. Since E is the inductive −1 limit space, there exists u = (Wn)n ∈ U such that Wn ⊂ 3(4) U for each n ∈ N. −1 −1 Hence Tu ⊂ 4 U.Takex ∈ Fu such that y ∈ x + 3 Tu. We need to prove that (x + Tu) ∩ P is a relative neighbohood of y ∈ P and (x + Tu) ∩ P ⊂ y + U. Set −1 −1 C := {x + 3 Tu : x ∈ Fu,y/∈ x + 3 Tu}.

Hence C (since Fu is ﬁnite) is closed and y/∈ C. Consequently, y ∈ P \ C is a relative open set, and −1 P \ C ⊂ y + 2(3) Tu ⊂ x + Tu. Finally, −1 −1 −1 −1 x + Tu = (x + 3 Tu) + 2(3) Tu ⊂ (y + 2(3) Tu) + 2(3) Tu = −1 y + 4(3) Tu ⊂ y + U.

Since every second-countable regular space is metrizable [146], every second- countable subset of a Hausdorff lcs is metrizable. Therefore we note the following result due to Cascales and Orihuela [81].

~~Theorem 10.2 (Cascales–Orihuela) Every precompact set in an (LM)-space E is metrizable. Hence E is angelic.~~

~~10.2 The case of dual metric spaces: elementary approach~~

Recall that a sequence (An)n of absolutely convex sets in an lcs E is called bornivorous if for every bounded set B in E there exists m ∈ N such that Am absorbs B (i.e., there exists t>0 such that B ⊂ tAm). Let us start with the following lemma.

Lemma 10.2 Let P be a precompact set and let (An)n be an increasing bornivorous sequence of absolutely convex sets in an ∞-quasibarrelled space E. Then there exists m ∈ N, a ﬁnite subset F ⊂ P such that P ⊂ F + Am.

Proof For elements x1,...,xk in P , k ∈ N,letTk be the absolutely convex hull of −2 {x1,...,xk}. Set Ck := 2Tk + 2 Ak. Assume that there exist x1,x2,...,xk ∈ P such that P ⊂ Ck. Since 2Tk is a compact subset of a finite-dimensional vector space, there exist m ∈ N, m ≥ k, and a finite subset W of 2Tk such that Ck ⊂ W + −1 2 Am. Consequently, P is covered by finitely many nonvoid sets of the form (y + −1 −1 2 Am) ∩ P .Ifz ∈ (y + 2 Am) ∩ P , we note that P is also covered by finitely many sets z + Am with z ∈ P . This proves that there exists a finite set F ⊂ P such that P ⊂ F + Am. 242 10 Precompact Sets in (LM)-Spaces and Dual Metric Spaces

Now assume that P Ck for each Ck of the form above. Inductively we select a sequence (xk)k in P such that xk+1 ∈/ Ck, k ∈ N. Since every Ck is absolutely convex and closed, for every k ∈ N there exists a continuous linear functional fk over E such that fk(xk+1)>1 and |fk(y)|≤1 for all −2 y ∈ 2Tk + 2 Ak.

~~Since (Ak)k is bornivorous, the sequence (fk)k is strongly bounded (i.e., bounded ∞ in β(E ,E)). The space E is -quasibarrelled, so (fk)k is equicontinuous. Fix 0~~

~~We are ready to prove the following result due to Valdivia [423,Note4].~~

~~Theorem 10.3 (Valdivia) Every precompact set in a dual metric space E is metrizable. Hence E is angelic.~~

Proof Let (Bn)n be a fundamental sequence of bounded absolutely convex sets in E.LetU be the family of all sequences u := (Un)n of the sets of the form Un := −1 k Bn, k,n ∈ N.Foru = (Un)n,set −1 An := 3 ac Uj . j≤n

Since the sequence (An)n of absolutely convex sets is increasing and bornivorous, we apply Lemma 10.2 to get a ﬁnite subset Fu ⊂ P and n(u) ∈ N such that −1 P ⊂ Fu + 3 ac Uj . j≤n(u) Set −1 Tu := 3 ac Uj . j≤n(u)

As the sets Bn are bounded, for each neighborhood of zero U in E there exists −1 u ∈ U such that Tu ⊂ 4 U. Then we follow similarly as in the proof of Theorem 10.1 to get the metrizability of the set P .

~~The last theorem yields the following result due to Pﬁster [329].~~

~~Theorem 10.4 (Pﬁster) Every precompact set in a (DF )-space E is metrizable. Hence E is angelic. Chapter 11 Metrizability of Compact Sets in the Class G~~

Abstract This chapter introduces (after Cascales and Orihuela) a large class of locally convex spaces under the name the class G. The class G contains among others all (LM)-spaces (hence (LF )-spaces), and dual metric spaces (hence (DF )- spaces), spaces of distributions D(Ω) and spaces A(Ω) of real analytic functions on open Ω ⊂ Rn. We show (following Cascales and Orihuela) that every precompact setinanlcsintheclassG is metrizable. This general result covers many already known theorems for (DF )-spaces, (LF )-spaces and dual metric spaces.

~~11.1 The class G: examples~~

Following Cascales and Orihuela [82], an lcs E is said to be in the class G if there is { : ∈ NN} a family Aαα of subsets of E (called a G-representation of E) such that N (a) E = {Aα : α ∈ N }, (b) Aα ⊂ Aβ if α ≤ β, and (c) in each Aα all sequences are equicontinuous. Condition (c) implies that every set Aα is σ(E ,E)-relatively countably compact. Therefore, if E is in the class G, the space (E,σ(E,E))has a relatively countably compact resolution. The class G is rich, containing (LM)-spaces, the dual metric spaces (hence (DF )-spaces), the space of distributions D(Ω) and the space A(Ω) of the real analytic functions for open Ω ⊂ RN, for example; see [89], [157]. The next proposition shows that the class G is stable by taking subspaces, separated quotients, completions, countable direct sums and countable products [82].

Proposition 11.1 (i) Let (En)n be a sequence of lcs’s in the class G. Then the := topological direct sum E n En belongs to the class G. (ii)Let (En)n be a sequence of lcs’s in the class G. Then the topological product := E n En belongs to the class G. (iii) If E is an lcs in the class G and F is a closed subspace, the quotient space E/F belongs to the class G. (iv) Every subspace F of an lcs E in the class G belongs to the class G. (v) The completion F of an lcs E in the class G belongs to the class G.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 243 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_11, © Springer Science+Business Media, LLC 2011 244 11 Metrizability of Compact Sets in the Class G

n n n n N Proof (i) Let {A n : α = (a ,a ,...,)∈ N } be a G-representation of E for each α 1 2 n n ∈ N A := Aj , α = (a1,a1,a2,a1,a2,a3,a1,a2,...,) . Set α j αj with 1 2 1 3 2 1 4 3 . Then the N family {Aα :α ∈ N } is a G-representation of E of sets in E (that is isomorphic to the product n En). { n : ∈ NN} ∈ N (ii) Let Aα α be a G-representation of En for each n . Set := 1 ⊕ 2 ⊕ a1 Aα Aα Aα ... Aα N N for α = (ak) ∈ N . Then {Aα: α ∈ N } is a G-representation for E in E (that is isomorphic to the direct sum En). N n (iii) Let {Aα : α ∈ N } be a G-representation of E. Then ⊥ N {Aα ∩ F : α ∈ N } is a G-representation of E/F in (E/F ), where F ⊥ is the orthogonal subspace to F in E. N (iv) If {Aα : α ∈ N } is a G-representation of E in E , N {Aα|F : α ∈ N } is a G-representation of F in F . (v) Since E and F have the same equicontinuous sets, the conclusion follows.

We provide short arguments showing that (DF )-spaces and (LM)-spaces E admit a G-representation. Let E be a (DF )-space.Let(Bn)n be a fundamental sequence of absolutely = ∈ NN := ◦ convex bounded subsets of E. For every α (nk) ,setAα k nk(Bk) . Clearly, the conditions (a), (b) and (c) are satisfied. Note also that for a bounded N subset B in (E ,β(E ,E)) there exists α ∈ N such that Aα absorbs B; such a G-representation will be called bornivorous. Let E be an (LM)-space.Let(Ej )j be a defining sequence for E of metrizable j lcs. For every j ∈ N,let(Un )n be a decreasing basis of absolutely convex neighborhoods of zero in Ej such that j j j U + + U + ⊂ Un n 1 n 1 j,n ∈ N α = (n ) ∈ NN A := (U k )◦. for all . For every k ,set α k nk Clearly, the conditions (a), (b) and (c) are satisfied. Since every bounded set B in (E,β(E,E)) is equicontinuous, the polar D of B is a neighborhood of zero in E. Hence there exists = NN k ⊂ ∈ N a sequence α (nk) in such that Un D for any k . Consequently, k B ⊂ D◦ ⊂ (U k )◦. nk k Applying Proposition 3.8, we have that for an uncountable-dimensional vector space E neither (E, σ (E, E∗)) nor (E∗,σ(E∗,E))is in the class G. This yields the following examples of spaces not in the class G.

Corollary 11.1 Let {El : i ∈ I} be an uncountable family of nonzero lcs. Then := := neither the topological direct sum S l El nor the topological product P l El is in the class G. 11.2 CascalesÐOrihuela theorem and applications 245

∈ := Proof Let Fl be a one-dimensional vector subspace of El for each l I . Then M ⊂ := ⊂ ∗ ∗ l Fl S and N l Fi P . Note that the weak dual of M is (M ,σ(M ,M)), and the weak∗ dual of N is (M, σ (M, M∗)). By the remark above, neither M nor N is in the class G. Therefore, neither S nor P is in the class G.

~~We also have the following proposition.~~

~~Proposition 11.2 If E is an lcs such that E is uncountable-dimensional, the space (E, σ (E, E)) is not in the class G.~~

Proof Assume (E, σ (E, E)) is in the class G. Since the completion of an lcs in the class G belongs to G, the completion (E∗,σ(E∗,E)) is in the class G.Thisis impossible by the remark before Corollary 11.1.

~~11.2 Cascales–Orihuela theorem and applications~~

~~Spaces in the class G enjoy another important general property, as follows.~~

~~Theorem 11.1 (CascalesÐOrihuela) Every precompact set in an lcs E in the class G is metrizable.~~

~~We already provided simple proofs for (LM)-spaces and (DF )-spaces. The following simple and short proof is due to Ferrando, K ˛akol and López-Pellicer [155].~~

{ : ∈ NN} = ∈ NN Proof Let Aα α be a G-representation of E.Forα (nk) ,set := { : = ∈ NN = ≤ ≤ } Cn1,...,nk Aβ β (mk) ,nj mj , 1 j k . By Dn1,n2,...,nk , we de- ∈ N note the polar of Cn1,n2,...,nk for each k .LetP be a precompact set in E. Since the completion of an lcs in the class G belongs to G, we may assume that P is compact. Note that for each ε>0 there is a countable subset Hε in E such that ◦ E = Hε + ε(P) . Indeed, otherwise (by the Zorn lemma) there exists an uncountable subset F in E, ε>0, such that the condition f − g ∈ ε(P)◦ for f,g ∈ F implies f = g.Byan obvious induction procedure, we select a sequence (nk)k in N and a sequence (fk)k ∈ in E of different elements with fk Cn1,n2,...,nk such that the condition ◦ fn − fm ∈ ε(P) = ∈ N ∩ implies m n. Indeed, there exists n1 such that F Cn1 is uncountable. Choose ∈ ∩ f1 F Cn1 . Since = { : ∈ N} Cn1 Cn1,m2 m2 , 246 11 Metrizability of Compact Sets in the Class G

∈ N \{ } ∩ there exists n2 such that (F f1 ) Cn1,n2 is uncountable. Select ∈ \{ } ∩ f2 (F f1 ) Cn1,n2 . ∈ Using a simple induction, we obtain both sequences as desired. As fk Cn1,n2,...,nk for all k ∈ N, the sequence (fk)k is equicontinuous. Indeed, for every k ∈ N, there = k ∈ NN ∈ = k = exists βk (mn)n such that fk Aβk , where nj mj for j 1, 2,...,k. Deﬁne = k : ∈ N an max mn k = ∈ NN ≥ ∈ N ⊂ and γ (an) . Note that γ βk for every k . Therefore Aβk Aγ , and so fk ∈ Aγ for all k ∈ N (by the condition (b)). Also, by (c) the sequence (fk)k is equicontinuous. Applying the Ascoli theorem for the Banach space Cc(P ),we obtain two different natural numbers j,k such that ◦ fj − fk ∈ ε(P) , which yields a contradiction. This proves the claim. Since := { : ∈ N} H Hn−1 n is countable, the topology τH on E of the pointwise convergence on H restricted to P is Hausdorff and metrizable, and coincides with the original topology of P . Hence P is metrizable.

For an lcs E,letτpc(E ,E)be the topology on E of the uniform convergence on precompact subsets of E. Note that if E is in the class G, the space (E,σ(E,E)) admits a relatively countably compact resolution; this resolution is also τpc(E ,E)- precompact (since every σ(E ,E)-relatively countably compact set is τpc(E ,E)- precompact). This may suggest the following question: Let E be an lcs whose dual E endowed with the topology τpc(E ,E) admits a precompact resolution. Does E belong to the class G? The answer is “no.” Indeed, let E be an inﬁnite-dimensional separable reﬂexive Banach space. Since (E,β(E,E))is a separable Banach space, it admits a compact resolution. Clearly, β(E ,E) is the topology τpc(E ,E) of the uniform convergence on σ(E,E)-precompact sets. (E, σ (E, E)) does not belong to the class G; this is a consequence of Proposition 11.2. Since precompact sets in an lcs in the class G are metrizable, each lcs in G is angelic. It turns out that the following stronger fact also holds; see [82, Theorem 11].

~~Proposition 11.3 The weak topology σ(E,E) of an lcs E in the class G is angelic.~~

Proof By the assumption, the space (E,σ(E,E))is web-compact. Applying The- orem 4.5, we derive that Cp(E ,σ(E ,E))is angelic. (E, σ (E, E )) is a topological subspace of Cp(E ,σ(E ,E)), and the conclusion follows.

We proved that every lcs in the class G is weakly angelic. This refers to many spaces, except the class of spaces Cp(X). Indeed, Proposition 12.2 below states that Cp(X), for uncountable spaces X, does not belong to the class G. Nevertheless, it 11.2 CascalesÐOrihuela theorem and applications 247 is known that Cp(X, E) is weakly angelic for any web-compact X and any lcs E in the class G;see[83, Theorem 8, Corollary 1.8]. Proposition 11.4 provides a direct proof of this fact.

Proposition 11.4 If X is a web-compact space and E is an lcs in the class G, the space Cp(X, E) is weakly angelic. If E ∈ G is separable and Cp(X) is angelic, the space Cp(X, Eσ ) is angelic, where Eσ := (E, σ (E, E )).

Proof Let {Aα : α ∈ Σ} be a web-compact representation for X. Set G := Cp(X, E). By Corollary 4.6, it is enough to show that (G ,σ(G,G)) contains a dense web-compact subset. If (gt )t → g in Cp(X, E), for each s ∈ X and each x ∈ E, then (x gt (s))t → x g(s). Therefore the map

δsx : Cp(X, E) → R deﬁned by δsx (g) := x g(s) is continuous. The set Z := δsx : s ∈ {Aα : α ∈ Σ},x ∈ E is a dense subset of (G ,σ(G,G)). Indeed, if f ∈ Cp(X, E), 0 = δsx (f ) = x f(s) for each s ∈ {Aα : α ∈ Σ}, and x ∈ E , then f(s)= 0 for each s ∈ {Aα : α ∈ Σ}. Then, by the continuity, f(s)= 0 for each s ∈ X. Hence f = 0. N Let E = {Bβ : β ∈ N } be a G-representation of E. Then N Z = {Dαβ : (α, β) ∈ Σ × N }, where Dαβ ={δsx : s ∈ Aα,x ∈ Bβ }. To show that Z is a web-compact subset of (G,σ(G,G)), we need to prove that, N if ((αn,βn))n → (α, β) in Σ × N and for each n ∈ N

~~∈ δsnxn Dαnβn , the sequence (δsnxn )n has an adherent point in (G ,σ(G,G)).As~~

((αn,βn))n → (α, β), { : ∈ N} { : we note that sn n is a relatively countably compact subset of X, and xn N n ∈ N} is an equicontinuous subset of Bγ , where γ is an element of N that veriﬁes βn ≤ γ for each n ∈ N . Then the sequence (δs x ) has a subnet n n n δ sn(d)xn(d) d∈D 248 11 Metrizability of Compact Sets in the Class G

→ ∈ such that (sn(d))d∈D s X and → (xn(d))d∈D x → in (E ,σ(E ,E)). From the equicontinuity, it follows that (xn(d))d∈D x uniformly on the precompact subsets of E. The proof will be ﬁnished if we show that

δ → δsx sn(d)xn(d) d∈D in (G,σ(G,G)). In other words, we have to prove that for each f ∈ C(X,E) we have [ ]= [ ] lim xn(d) f(sn(d)) x f(s) . d∈D This equality follows from the following facts: [ ]= [ ] (*) limd∈D x f(sn(d)) x f(s) . (**) As {f(sn) : n ∈ N} is a relatively countably compact subset of E (therefore precompact), [ ]= [ ] lim xn(d) f(sn(d)) lim x f(sn(d)) . d∈D d∈D Now assume that E is separable. Then (E,σ(E,E)) is separable; see Corol- lary 12.5 below. If G is a countable and dense subset in (E,σ(E,E)), ξ := σ(E,G) is a metrizable locally convex topology on E with ξ ≤ σ(E,E). The assumptions of Theorem 4.3 are satisﬁed: Cp(X) is angelic and Eξ is metrizable, so Cp(X, Eξ ) is angelic, where Eξ := (E, ξ). Note that Cp(X, Eσ ) ⊂ Cp(X, Eξ ). Then we deduce that Cp(X, Eσ ) is angelic.

~~Applying Proposition 11.4, we provide the following vector-valued version of Theorem 9.9.~~

Theorem 11.2 Let E be a separable lcs in the class G. Let ξ be a regular topology on C(X,E) stronger than the pointwise topology of C(X,E). The following assertions are equivalent: (i) (C(X, E), ξ) is K-analytic. (ii) (C(X, E), ξ) admits a compact resolution. (iii) (C(X, E), ξ) admits a relatively countably compact resolution.

Proof Since each K-analytic space admits a compact resolution, it is enough to show (iii) ⇒ (i): If (C(X, E), ξ) admits a relatively countably compact resolution N N {Kα : α ∈ N }, the family {Kα : α ∈ N } is a bounded resolution on Cp(X, E) in the topology τp. Cp(X) is isomorphic to a subspace of Cp(X, E). Hence Cp(X) admits a bounded resolution. By Theorem 9.15(i), the space υX is web-compact, so Cp(υX) is angelic by Proposition 4.2. Consequently, Cp(X) is angelic by Lemma 9.2. Now we apply Proposition 11.4 to conclude that Cp(X, Eσ ) is angelic. As Cp(X, E) ⊂ Cp(X, Eσ ), the space Cp(X, E) is angelic, and then by The- orem 4.1 the space (C(X, E), ξ) is also angelic. Finally, since angelic spaces having 11.2 CascalesÐOrihuela theorem and applications 249 a resolution consisting of relatively countably compact sets are K-analytic (Corol- lary 3.6), the space (C(X, E), ξ) is K-analytic.

~~We present a couple of examples (adopted from [78]) motivated by Theo- rem 11.2.~~

~~Example 11.1 If X is an Eberlein compact space and Y is a Polish space, Cp(X, Y ) is an angelic space having a compact resolution. Hence Cp(X, Y ) is K-analytic.~~

N Proof Since every Polish space is a Gδ-subset of R , it is enough to show that N Cp(X, Y ) has a compact resolution if Y is an open subset of R . Since N N Cp(X, R ) = (Cp(X)) N N and X is Eberlein compact, Cp(X, R ) has a compact resolution {Kα : α ∈ N }.Let ρ be a metric deﬁning the topology of RN. For an open subset Y of RN, the family N {Kα ∩ Cp(X, Y ) : α ∈ N } is a resolution on Cp(X, Y ).Ifα = (an)n, then := { ∈ ∩ : RN\ ≥ −1 ∈ } Hα f Kα Cp(X, Y ) ρ(f(x), Y) a1 ,x X is a closed subset of Kα and hence compact. If f ∈ Cp(X, Y ), then f(X)⊂ Y is compact, and there exists p ∈ N such that ρ(f(x),RN\Y)≥ p−1 for each x ∈ X. There exists β = (bn)n such that f ∈ Kβ . Deﬁne α = (an)n by N a1 = b1 + p, an = bn,n 2. Then f ∈ Hα. Hence {Hα : α ∈ N } is a compact N resolution of Cp(X, Y ). By Proposition 11.4, the space Cp(X, R ) is angelic, so its subset Cp(X, Y ) is angelic, too. Finally, we deduce that Cp(X, Y ) is K-analytic.

Example 11.2 If X is a separable and normed space and E is a Fréchet–Montel space, the space Lc(X, E) of all continuous linear maps from X into E endowed with the compact-open topology is K-analytic.

Proof Let B be the unit ball in X, and let {xn : n ∈ N} be a countable dense subset of E (every Fréchet–Montel space is separable; see Corollary 6.6). Set ∞ nk −1 Kα := B(xj ,k ), k=1 j=1 −1 where B(xj ,k ) is the closed ball in E with its center at the point xj and radius −1 N N k for α = (nk) ∈ N and all j,k ∈ N. Then {Kα : α ∈ N } is a compact resolution on E, and each compact set K in E is contained in some Kα. Set

Dα := {f ∈ L(X, E) : f(B)⊂ Kα}. N By Ascoli’s theorem, the family {Dα : α ∈ N } is a relatively compact resolution on Lc(X, E). We apply Proposition 11.4 to show that Cp(X, Eσ ) is angelic. The inclusion Cp(X, E) ⊂ Cp(X, Eσ ) and Theorem 4.1 imply that Cc(X, E) is angelic, too. Then Lc(X, E) ⊂ Cc(X, E) is angelic. Corollary 3.6 shows that Lc(X, E) is K-analytic. 250 11 Metrizability of Compact Sets in the Class G

~~2 2 Example 11.3 If I is uncountable, Lp( (I), (N)) does not admit a compact 2 2 resolution and Lc( (I), (N)σ ) admits a compact resolution.~~

2 2 Proof Since Cp( (I)σ , (N)σ ) is angelic (by the second part of Proposition 11.4), 2 2 we note that the space Lp( (I), (N)) is angelic, as it is isomorphic to the an- 2 2 2 2 gelic space Lp( (I)σ , (N)σ ).From[389], the space Lp( (I), (N)) is not 2 2 K-analytic, so Lp( (I), (N)) does not have a compact resolution (since the last space is angelic). By Theorem 3.5, there exists on 2(N) a quasi-(LB)- N 2 representation {Aα : α ∈ N } such that every Banach disc in (N) is contained 2 in some Aα. For each α ∈ N,letBα be the closure of Aα in (N). Set 2 2 Kα := {f ∈ L( (I), (N)σ ) : f(B)⊂ Bα}, where B is the unit ball in 2(I). Applying the Ascoli theorem, we deduce that N 2 2 {Kα : α ∈ N } is a resolution on Lc( (I), (N)σ ) of relatively compact sets; hence 2 2 Lc( (I), (N)σ ) admits a compact resolution.

A similar argument yields the following: If E is a separable normed space and F is a reﬂexive Fréchet space, Lc(E, Fσ ) is K-analytic. By Proposition 11.4, the space 2 Cp(X, C[0, 1]σ ) is angelic, where X is the closed unit ball in (I) endowed with the weak topology. Canela [78] proved that, if I is uncountable, Cp(X, C[0, 1]σ ) is not K-analytic. Hence we have the following example.

Example 11.4 Let X be the closed unit ball in 2(I) endowed with the weak topology for uncountable I . Then Cp(X, C[0, 1]σ ) does not admit a compact resolution. Chapter 12 Weakly Realcompact Locally Convex Spaces

Abstract In this chapter, we continue the study of spaces in the class G. We prove that the weak∗ dual (E,σ(E,E)) of an lcs E in the class G is K-analytic if and only if (E,σ(E,E)) is Lindelöf if and only if (E, σ (E, E)) has countable tightness. We show that every quasibarrelled space in the class G has countable tightness both for the weak and the original topologies. This extends a classical result of Kaplansky for a metrizable lcs. Although (DF )-spaces belong to the class G, concrete examples of (DF )-spaces without countable tightness are provided. On the other hand, there are many Banach spaces E for which E endowed with the weak topology is not Lindelöf. We show, however (following Khurana), that every WCG Fréchet space E is weakly K-analytic. An example due to Pol showing that there exists a Banach space C(X) over a compact scattered space X such that C(X) is weakly Lindelöf and not WCG is presented. We show (after Amir and Linden- strauss) that every nonseparable reﬂexive Banach space contains a complemented separable subspace. Several consequences are provided.

~~12.1 Tightness and quasi-Suslin weak duals~~

This section deals with an lcs E whose weak∗ dual (E,σ(E,E))is K-analytic (or analytic, or at least quasi-Suslin). A classical result of Kaplansky (see [165, Theorem, p. 37]) states that if X is a σ -compact space and Z is a metric space, the space Cp(X, Z) has countable tightness. This is applied to show that the weak topology of a metrizable lcs E has countable tightness. Indeed, since E is metrizable, (E,σ(E,E))is σ -compact, so Cp(E ,σ(E ,E))has countable tightness by Kaplansky’s result. On the other hand, as (E, σ (E, E )) ⊂ Cp(E ,σ(E , E)), we deduce that the space (E, σ (E, E)) also has countable tightness. Recall that, for an lcs E,byτpc(E ,E)we denote the topology on E of the uniform convergence on precompact sets of E. Clearly, σ(E ,E)≤ τpc(E ,E), and if E is quasicomplete, then also τpc(E ,E)≤ μ(E ,E).Although the Mackey dual of an analytic space E need not be analytic (see Theorem 6.6), we know from Propo- sition 6.17 that (E ,τpc(E ,E)) is analytic for each separable (LF )-space E.We

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 251 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_12, © Springer Science+Business Media, LLC 2011 252 12 Weakly Realcompact Locally Convex Spaces show that the weak∗ dual of any lcs in the class G is a quasi-Suslin space; see The- orem 12.1. Valdivia [421] proved that, if E is a Fréchet space and F is the strong dual of E, the space (F ,σ(F,F))is quasi-Suslin, and (F ,σ(F,F))is K-analytic if and only if (F, μ(F, F )) is barrelled. Clearly, F ,asa(DF )-space, has a closed G- N representation {Kα : α ∈ N } (i.e., each set Kα is σ(F ,F)-closed). This motivates Theorem 12.1 (proved in [159, Theorem 4]) which extends [80, Proposition 1]. We say that a subset A ⊂ E is full if it contains all adherent points in E of sequences from A.

~~Theorem 12.1 Let E be an lcs in the class G. Then (E,σ(E,E))is a quasi-Suslin space.~~

~~N N Proof Let {Aα : α ∈ N } be a G-representation for E. For each α ∈ N , deﬁne ◦◦ Bα := {S : S ⊂ Aα, |S|≤ℵ0}.~~

As countable unions of countable sets are countable, each sequence (un)n in Bα belongs to a bipolar of a sequence (vn)n in Aα that is clearly equicontinuous by N the assumptions on {Aα : α ∈ N }. By the Alaoglu–Bourbakiˇ theorem (see [213, ◦◦ Theorem 8.5.2]), the set {vn : n ∈ N} is absolutely convex, equicontinuous and σ(E ,E)-compact. Hence Bα is absolutely convex and (un)n has cluster points that ◦◦ belong to {vn : n ∈ N} ⊂ Bα. This proves that Bα is absolutely convex, weakly N countably compact and full. Also, the family {Bα : α ∈ N } is a G-representation of E. From Proposition 3.11, it follows that E is quasi-Suslin.

~~This implies the following corollary; see [82].~~

Corollary 12.1 Let E be an lcs in the class G such that (E, σ (E, E)) is web- compact. Then (E,σ(E,E)) is K-analytic. In particular, every separable lcs in class G has its precompact dual (E ,τpc) analytic.

Proof Since (E ,σ(E ,E))is a subspace of Cp(E, σ (E, E )) (the last space is angelic by Theorem 4.4), the space (E,σ(E,E)) is angelic. By Theorem 12.1,the space (E,σ(E,E)) is quasi-Suslin. Apply Corollary 3.6. For the particular case, see Proposition 6.3.

~~Corollary 12.1 extends Proposition 6.17 since (LF )-spaces belong to the class G. We need the following characterization of a weakly realcompact lcs due to Corson, see [421, p. 137] for the proof.~~

Proposition 12.1 Let (E, E ) be a dual pair. Let {Fi : i ∈ I} be the family of all separable closed subspaces of (E,σ(E,E)). Then the following statements are equivalent: (i) (E, σ (E, E)) is realcompact. ∗ (ii) E ={z ∈ (E ) : z|Fi is σ(E ,E)-continuous for each i ∈ I}. 12.1 Tightness and quasi-Suslin weak duals 253

~~We are ready to prove the following important result [88] for which the implication (i) ⇒ (iv) is true in general; see Lemma 9.6.~~

Theorem 12.2 (Cascales–Kakol–Saxon) ˛ Let E be an lcs in the class G. The following statements are equivalent: (i) (E, σ (E, E)) has countable tightness. (ii) For each space Y , afunctionfromE into Y that is σ(E,E)-continuous restricted to σ(E,E)-closed and separable subsets of E is σ(E,E)-continuous on E. (iii) A linear functional on E is σ(E,E)-continuous whenever its restriction to every σ(E,E)-closed and separable subspace of E is σ(E,E)-continuous. (iv) (E,σ(E,E))is realcompact. (v) (E,σ(E,E))is K-analytic. (vi) (E,σ(E,E))n is Lindelöf for every n ∈ N. (vii) (E,σ(E,E))is Lindelöf.

Proof (i)⇒(ii): Let f : E → Y beamapσ(E,E)-continuous when restricted to the σ(E,E)-closed and separable subsets of E. It is enough to show that for any σ(E,E) set A ⊂ E and x ∈ A we have f(x)∈ f(A). By the assumption, there is σ(E,E) countable D ⊂ A such that x ∈ D , and thus f | σ(E,E) is continuous. Hence D f(x)∈ f(D)⊂ f(A). (ii)⇒(iii) is obvious. (iii)⇒(iv): This follows from Proposition 12.1. (iv)⇒(v): By Theorem 12.1, the space (E,σ(E,E))is quasi-Suslin, so by The- orem 3.1 (E,σ(E,E))admits a resolution of relatively countably compact sets. By the assumption, (E,σ(E,E)) is realcompact, so every relatively countably compact set is relatively compact. Now it is enough to apply Proposition 3.13 (v) to deduce that (E,σ(E,E))is K-analytic. (v)⇒(vi): Since countable products of K-analytic spaces are K-analytic, and K- analytic spaces are Lindelöf, the conclusion follows. (vi)⇒(vii) is clear. (vi)⇒(i): Since (E,σ(E,E))n is a Lindelöf space for all n ∈ N, by Propo- sition 9.9, the space Cp(E ,σ(E ,E)) has countable tightness. Since subspaces of spaces with countable tightness have countable tightness, we note that (E, σ (E, E)) has countable tightness. Finally, since Lindelöf spaces are realcompact (Proposition 3.12), we have (vii)⇒(iv).

There is another approach to proving Theorem 12.2 by using Proposition 9.15. Indeed, if E ∈ G, then by Theorem 12.1 the space (E,σ(E,E)) is quasi-Suslin. Consequently, the space υ(E,σ(E,E)) is K-analytic by Proposition 9.15. Since the countable tightness of (E, σ (E.E)) of an lcs E implies that (E,σ(E,E)) is realcompact (by applying Lemma 9.6), the space (E,σ(E,E)) is a K-analytic space if E ∈ G and (E, σ (E.E)) has countable tightness. 254 12 Weakly Realcompact Locally Convex Spaces

A topological space X is said to have the countable Suslin number (c(X) ≤ℵ0) if the cardinality of every pairwise disjoint family of open sets in X does not exceed ℵ0. It is known that every regular Lindelöf space is a paracompact space; this is a result due to Morita [303]. Moreover, every paracompact space having the countable Suslin number is a Lindelöf space; see [28, Chapter 2, Exercise 393]. By [47] and [341] the weak topology σ(E,E) of a Banach space E is Lindelöf if and only if σ(E,E) is paracompact if and only if σ(E,E) is normal. We note also the following fact that supplements Theorem 12.2.

~~Corollary 12.2 Let E be an lcs. Then (E, σ (E, E)) (resp. (E,σ(E, E))) is a Lindelöf space if and only if (E, σ (E, E)) (resp. (E,σ(E, E))) is a paracompact space.~~

Proof Every regular Lindelöf space is paracompact. Now assume that the space (E,σ(E,E))is paracompact. Since (E,σ(E,E))is dense in the product RI for some set I , and the Suslin number c(RI ) of RI is countable for any set I (see, for example, [28, Chapter 2, Exercise 383]), we note that c(E ,σ(E ,E))=ℵ0. Then, by the result mentioned in [28, Chapter 2, Exercise 393], we have that (E,σ(E,E)) E is Lindelöf. Since (E, σ (E, E )) ⊂ Cp(E ,σ(E ,E)) ⊂ R , the remaining case concerning (E, σ (E, E)) is proven similarly.

~~12.2 A Kaplansky-type theorem about tightness~~

We provide a large subclass in G of lcs having countable tightness. The following theorem due to Cascales, Kakol ˛ and Saxon [88, Proposition 4.8] extends Kaplan- sky’s result [165] stating that the weak topology of a metrizable lcs has countable tightness. Recall again that E is (quasi)barrelled if every σ(E,E)-bounded (every β(E,E)-bounded) set is equicontinuous. Every (LM)-space (every (LF )-space) is quasibarrelled (barrelled).

~~Theorem 12.3 (Cascales–Kakol–Saxon) ˛ Every quasibarrelled space E in G has countable tightness, and the same also holds true for (E, σ (E, E)).~~

N Proof Let {Aα : α ∈ N } be a G-representation of E. Since E is quasibarrelled and condition (c) holds, each Aα is equicontinuous. Replacing each Aα by its σ(E ,E)- closed, absolutely convex hull, we may assume that each Aα is a β(E ,E)-Banach disc (the strong dual of quasibarrelled spaces must be quasicomplete). This implies that the space (E,β(E,E))is a quasi-(LB)-space, and therefore, using The- orem 3.5, there exists a family of β(E,E)-Banach discs of E (which we again N denote by {Aα : α ∈ N }) such that: N (i) E =∪{Aα : α ∈ N }. N (ii) Aα ⊂ Aβ if α ≤ β in N . 12.2 A Kaplansky-type theorem about tightness 255

N (iii) For every β(E ,E)-Banach disc B ⊂ E , there is α ∈ N such that B ⊂ Aα. W ={ } Consider a web Cn1,n2,...,nk , where each Cn1,n2,...,nk is defined as usual. Then W is a web having the following properties: Cn ,n ,...,n ⊂ Cm ,m ,...,m for nj ≤ mj if k ∈ N and 1 ≤ j ≤ k. 1 2 k 1 2 kN For every α = (nk)k ∈ N and every β(E ,E)-neighborhood of zero U ⊂ E , ∈ N ≥ ⊂ there exist nU , pU 0, such that Cn1,n2,...,nU pU U. The order condition follows from the definitions. The remaining condition we check as follows: Every set Aα (being a Banach disc) is β(E ,E)-bounded. Note that the web W is bounded. Indeed, assume that this does not hold. Then we find N α = (nk)k ∈ N and a β(E ,E)-neighborhood U of 0 in E such that ⊂ Cn1,n2,...,nk kU ∈ N = k ∈ NN | = for all k . For every positive integer k, there is αk (an)n with αk k ⊂ (n1,n2,...,nk) such that Aαk kU. We now define = { k : ∈ N} an max an k for all n ∈ N and γ = (an)n. It is clear that γ ≥ αk and Aγ ⊂ kU if k ∈ N, which contradicts the boundedness of Aγ . Given positive integers k,n1,n2,...,nk, we define

:= σ(E ,E) Dn1,n2,...,nk Cn1,n2,...,nk . Since β(E,E)has a basis of neighborhoods of zero consisting of σ(E,E)-closed N sets, and as the web W is bounded, for every α = (nk)k ∈ N and every β(E ,E)- ∈ N ≥ ⊂ neighborhood U of zero in E there exist nU , pU 0, such that Dn1,n2,...,nU pU U. If we relabel ∞ := Aα Dn1,n2,...,nk , k=1 N then the new family {Aα : α ∈ N } still satisﬁes the desired properties. Since (E,β(E,E)) is quasicomplete, every β(E,E)-bounded set is contained in a N β(E ,E)-Banach disc, which means that the family {Aα : α ∈ N } is a fundamental family of equicontinuous subsets of E. Taking polars in the dual pair (E, E),we { ◦ : ∈ NN} note that the family Aα α is a basis of neighborhoods of zero in E.Onthe ⊂ = ∈ NN other hand, since Dn1,n2,...,nU pU U, it follows that for every α (nk)k the increasing sequence

~~D◦ ⊂ D◦ ⊂···⊂D◦ ⊂··· n1 n1,n2 n1,n2,...,nk is bornivorous. Then, by Proposition 2.13, we have for every ε>0 the inclusion~~

~~∞ σ(E,E ) ∞ ◦ ◦ A = D◦ ⊂ (1 + ε) D . (12.1) α n1,n2,...,nk n1,n2,...,nk k=1 k=1 256 12 Weakly Realcompact Locally Convex Spaces~~

~~N Summarizing all facts, we deduce that if we deﬁne for α = (nk)k ∈ N ∞ U := D◦ , α n1,n2,...,nk k=1~~

N the family {Uα : α ∈ N } is a basis of neighborhoods of zero in E. We need to show that the tightness of E is countable. Take an arbitrary set A ⊂ E ∈ ∈ with 0 A. Then the set B of elements xn1,n2,...,nk is countable and 0 B, where x is a chosen point in D◦ ∩ A, if n1,n2,...,nk n1,n2,...,nk

~~D◦ ∩ A = ∅,k,n,n ,...,n ∈ N. n1,n2,...,nk 1 2 k~~

Finally, we prove that (E, σ (E, E)) has countable tightness. Similar to what we did in the proof of (i)⇒(ii)⇒(iii) in Theorem 12.2, every linear functional on E that is continuous on each separable and closed subspace of E is continuous. The families of closed and separable subspaces of E and σ(E,E)-closed and separable subspaces of E, respectively, coincide. Hence the countable tightness of E implies the condition (iii) from Theorem 12.2, and so (E, σ (E, E)) has countable tightness.

~~Theorem 12.2 fails if E is not in the class G. To prove this, we need the following additional fact.~~

~~Proposition 12.2 The space Cp(X) belongs to the class G if and only if X is countable.~~

X Proof Since Cp(X) is a dense subspace of the product R , we apply Proposi- X tion 3.8. For another proof, assume Cp(X) belongs to the class G. Then R is a space in the class G (which clearly is a Baire space). Now Theorem 12.3 applies (since RX has countable tightness if and only if X is countable).

~~Example 12.1 There exists an lcs for which its weak topology has countable tightness and its weak∗ dual is not K-analytic.~~

Proof Let X be an uncountable regular Lindelöf P -space (i.e., every Gδ-set in X n is open). Since X is a Lindelöf space for any n ∈ N, the space Cp(X) has countable tightness (Theorem 9.9). By Proposition 12.2, the space Cp(X) is not in the := class G. Assume that F Cp(X)σ is K-analytic. Then F has a compact resolution N {Aα : α ∈ N } in F . Since X is a normal P -space (see Lemma 6.1), every topologically bounded set in X is ﬁnite, and by Proposition 2.15 the space Cp(X) = Cc(X) N is barrelled. Hence every set Aα is equicontinuous, so {Aα : α ∈ N } is a G- representation. Consequently, Cp(X) belongs to G, a contradiction.

~~Example 12.2 (Cascales–Kakol–Saxon ˛ ) There exist (DF )-spaces with uncountable tightness whose weak topology has countable tightness. 12.2 A Kaplansky-type theorem about tightness 257~~

Proof Let Λ be an uncountable indexing set. For each S ⊂ Λ, deﬁne ES ={u ∈ 2(Λ) : u(x) = 0forx/∈ S}. Let E be the Banach space 2(Λ) endowed with the coarsest topology ξ such that the projection of E onto the Banach space ES along EΛ\S is continuous for every countable S ⊂ Λ. A basis of neighborhoods of zero for E consists of the sets U of the form

U = V + EΛ\S, where V is a positive multiple of the unit ball in the Banach space 2(Λ) and S is a countable subset of Λ. The space (E, τ) := 2(Λ) admits the (DF )-space topology ξ that is not quasibarrelled, σ(E,E)<ξ<τ, and such that (E, ξ) does not have countable tightness. Note that σ(E,E)= σ(E,E) is K-analytic (since 2(Λ) is reﬂexive) and not analytic (as it is nonseparable). Since (E, τ) is metrizable, the topology σ(E,E) has countable tightness. On the other hand, the space (E, ξ) has uncountable tightness. Indeed, the set B of the characteristic functions of the singleton subsets of Λ has 0 in its closure but not in the closure of any countable subset of B. We see from this example that a (DF )-space may fail to have countable tightness even when the weak topology does.

Every WCG Banach space E is weakly K-analytic. We prove this result in The- orem 12.8. In particular, every reﬂexive Banach space E is weakly K-analytic. This follows also from Theorem 12.2. Indeed, for the strong dual (E,β(E,E)), Theo- rem 12.3 is applied to derive that σ(E,E) has countable tightness. (E, σ (E, E)) is K-analytic by Theorem 12.2. N Let E be an lcs in the class G. We call a G-representation {Aα : α ∈ N } bornivorous if every β(E ,E)-bounded set is contained in some Aα. It is easy to see that every dual metric space (hence any (DF )-space), every (LM)-space and every quasibarrelled space in the class G has a bornivorous G-representation; see [89, Lemma 2], [157, Propositions 1 and 2], [159, Theorem 9]. A general fact will be proved.

Theorem 12.4 (Cascales–Kakol–Saxon) ˛ Let E be an lcs in the class G. Let N {Aα : α ∈ N } be a bornivorous G-representation. The following assertions are equivalent: (i) The space (E, σ (E, E)) has countable tightness. (ii) The space (E, μ(E, E)) is quasibarrelled. (iii) The space (E, μ(E, E)) has countable tightness.

Proof (i) ⇒ (ii): Theorem 12.2 implies that (E,σ(E,E))is realcompact, and thus ◦◦ so is the closed subset Aα . The latter is also countably compact due to the borniv- ◦◦ ⊂ orous condition (which implies Aα Aβ ) and the equicontinuity of sequences (in ◦◦ ◦◦◦ = ◦ Aβ ). Hence Aα is σ(E ,E)-compact, and Aα Aα is a neighborhood of zero in the Mackey topology. Each β(E ,E)-bounded set is contained in some Aα and thus is μ(E, E)-equicontinuous. This proves that the Mackey topology is quasibarrelled. 258 12 Weakly Realcompact Locally Convex Spaces

N (ii) ⇒ (iii): Clearly, {Aα : α ∈ N } is a G-representation for (E, μ(E, E )) as well, and now (iii) follows from Theorem 12.3. (iii) ⇒ (i): Countable tightness of E := (E, μ(E, E)) implies condition (iii) from Theorem 12.2 (see the proof of Theorem 12.3, second part). By Theorem 12.2, the space (E, σ (E, E)) has countable tightness.

If [X, 1]:={f ∈ C(X) :|f(x)| 1 for all x ∈ X} is absorbing in C(X) (i.e., X is pseudocompact), by Cu(X) we denote the space C(X) endowed with the uniform norm topology having the unit ball [X, 1]. It is well known that if X is pseudocompact and Cc(X) = Cu(X) , the space X is compact.

Theorem 12.5 The following assertions are equivalent for a (df )-space Cc(X): (1) X is compact. (2) Cc (X) coincides with the Banach space Cu(X). (3) Cc (X) has countable tightness. (4) The weak topology of Cc (X) has countable tightness. (5) The Mackey topology μ(Cc (X) ,Cc (X) ) has countable tightness.

Proof (1) ⇒ (2) ⇒ (3)isobvious. (3) ⇒ (4) follows from the proof in Theorem 12.4 (iii) ⇒ (i) with E := Cc(X). (4) ⇔ (5) from Theorem 12.4. (5) ⇒ (1): By Theorem 12.4, condition (5) implies that the Mackey topology μ(Cc(X), Cc(X) ) is quasibarrelled, and therefore the bornivorous barrel [X, 1] (see Theorem 2.13)isaμ(Cc(X), Cc(X) )-neighborhood of zero. Hence Cc (X) = Cu(X) , which implies that X is compact.

~~Example 2.4 provided examples of (df )-spaces Cc(X) that are not (DF )-spaces. Thus, the following corollary provides a class of spaces that are quasi-Suslin and not K-analytic.~~

~~Corollary 12.3 Let E := Cc (X) be any (df )-space that is not a (DF)-space. Then the weak∗ dual of E is quasi-Suslin and not K-analytic.~~

~~Proof By Theorem 12.1, (E,σ(E,E))is quasi-Suslin. By Theorem 12.5 and The- orem 12.2, the space X is compact if and only if (E,σ(E,E))is K-analytic.~~

~~12.3 K-analytic spaces in the class G~~

Canela [78, Proposition 7] proved that a weakly K-analytic lcs E satisﬁes dens(E) ≤ dens(E,σ(E,E)). Moreover, if E is additionally metrizable, the equality dens(E) = dens(E,σ(E,E)) holds, where as usual dens(E) denotes the density of E. This result extended Talagrand’s [388, Theorem 6.1] statement of the same for WCG Banach spaces. Cascales and Orihuela [83, Theorem 13] extended Canela’s result to weakly Lindelöf Σ-spaces in the class G. Our next result extends the results above. 12.3 K-analytic spaces in the class G 259

~~Proposition 12.3 Let E be an lcs such that the weak∗ dual (E,σ(E,E)) is a quasi-Suslin space and (E, σ (E, E)) is a Lindelöf Σ-space. Then we have~~

~~dens(E ,σ(E ,E))= dens(E).~~

Proof Since (E, σ (E, E )) is a Lindelöf Σ-space, the space Cp(E, σ (E, E )) is angelic by applying Theorem 4.4.As(E ,σ(E ,E)) is included in Cp(E, σ (E, E )), we note that (E,σ(E,E)) is angelic. Now Corollary 3.6 is applied to conclude that (E,σ(E,E)) is K-analytic. Let B be a dense subset of E of the cardinality at most ℵ. Then σ(E,B) is Hausdorff, σ(E,B)≤ σ(E,E), and (E,σ(E,B)) has a basis of neighborhoods of zero of the cardinality at most ℵ. (E,σ(E,E)) is Lindelöf. Hence the space (E,σ(E,B))has a basis of open sets of the cardinality at most ℵ. By Lemma 3.1, we have dens(E,σ(E,E))≤ℵ. If B is a dense subset in (E,σ(E,E)) with |B|≤ℵ, the Lindelöf property yields ω(E,σ(E,B)) ≤ℵ. We apply Lemma 3.1 and conclude dens(E, σ (E, E)) ≤ℵ. From the equality dens(E, σ (E, E)) = dens(E), it follows that dens(E) ≤ℵ.

~~Corollary 12.4 Let E be an lcs in the class G such that (E, σ (E, E)) is a Lindelöf Σ-space. Then dens(E,σ(E,E))= dens(E).~~

~~Corollary 12.5 If E is a separable lcs such that (E,σ(E,E))is quasi-Suslin, the space (E,σ(E,E)) is analytic. In particular, then the weak∗ dual of a separable lcs in the class G is separable.~~

Proof The separability of E combined with Corollary 4.2 yields that the space (E,σ(E,E)) is angelic. Consequently, it is K-analytic by Corollary 3.6. Finally, by Proposition 6.3, we know that (E,σ(E,E))is analytic.

If for a compact space X the space C(X) is a WCG Banach space, X is Talagrand compact. According to a theorem of Amir and Lindenstrauss ([5];seealso[149, Theorem 12.12], [312], and Theorem 9.1), for a compact space X, the space C(X) is a WCG Banach space if and only if X is Eberlein compact. The following simple observation motivates Theorem 12.6.

~~Proposition 12.4 Let E be a metrizable lcs. Then every compact set in (E, σ (E, E)) is Talagrand compact.~~

Proof Since E is metrizable, (E,σ(E,E))is σ -compact. Hence (E,σ(E,E))is N web-compact with Σ = N . By Corollary 4.4, every compact set in Cp(E ,σ(E ,E)) is Talagrand compact. Finally, (E, σ (E, E )) ⊂ Cp(E ,σ(E ,E)), so the conclusion follows.

It turns out (see [82, Theorem 12]) that Talagrand compact sets can be characterized as weakly compact sets in an lcs in the class G. The implication (ii) ⇒ (i) in Theorem 12.6 follows also from Corollary 4.6. 260 12 Weakly Realcompact Locally Convex Spaces

Theorem 12.6 For a compact space X, the following conditions are equivalent: (i) X is Talagrand compact. (ii) There exists an lcs E in the class G such that X is homeomorphic to a σ(E,E)-compact subset of E.

Proof (ii) ⇒ (i): Let X be a σ(E,E)-compact subset of an lcs E in the class G. N N Let {Kα : α ∈ N } be a G-representation of E.IfAα := {f |X : f ∈ Kα},α∈ N , then every Aα is a subset of C(X). Moreover, Aα ⊂ Aβ if α ≤ β. Also, since Aα is relatively countably compact in Cp(X), it is relatively compact in Cp(X) (as = ⊂ Cp(X) is angelic). Set S α Aα. Since S Cp(X) separates points of X,we apply Corollary 3.6 and Theorem 9.2 (ii) to ensure that Cp(X) is K-analytic. Hence X is Talagrand compact. (i) ⇒ (ii): Assume X is Talagrand compact. Then the Banach space C(X) is N weakly K-analytic, so C(X) admits a compact resolution {Kα : α ∈ N } in the weak topology; see Theorem 9.2 (i). We may assume that the sets Kα are absolutely convex (by using the Krein theorem; see [213]). Let (C(X),ξ) be the dual of C(X) endowed with the topology of the uniform convergence on the sets Kα. Then (C(X),ξ) belongs to G, and the dual (C(X),ξ) equals C(X). Note that X is weakly compact in (C(X),ξ).

~~Proposition 12.5 Let Cp(X) be a separable web-bounded space with countable tightness. Then Lp(X) is separable.~~

Proof Since Cp(X) has countable tightness, X is realcompact by Proposition 9.9. The space Lp(X) is a Lindelöf Σ-space by Theorem 9.15.Letσ be the original topology of Lp(X). Since Cp(X) is separable, Lp(X) admits a weaker metrizable topology ξ ≤ σ . As the weight of ξ is countable, by Lemma 3.1 the density of σ is countable. Hence Lp(X) is separable.

~~The weak∗ dual of a separable lcs in the class G is a separable space by Corol- lary 12.5. This and Proposition 12.5 motivate the following problem.~~

~~Problem 12.1 Let E be a separable web-bounded lcs. Is the weak∗ dual of E separable?~~

~~12.4 Every WCG Fréchet space is weakly K-analytic~~

We know already (Theorem 12.2, Theorem 12.3) that the weak∗ dual of a quasibarrelled lcs in the class G is K-analytic. In particular, every reﬂexive Fréchet space is weakly K-analytic. In this section, we prove that every weakly compactly generated Fréchet space E is weakly K-analytic; this result is due to Khu- rana [242]. For the same result for WCG Banach spaces, see [390] and also [149] and [323]. 12.4 Every WCG Fréchet space is weakly K-analytic 261

A Banach space E is weakly compactly generated (WCG) if there exists a weakly compact subset K in E whose linear span is a dense subspace of E. One may assume that K is absolutely convex by the Krein theorem. Orihuela [323] used the method of constructing projections in WCG Banach spaces (this method is due to Valdivia [426], [428], [429], [430], [431] and [321]) to provide a direct proof that the weak topology of a WCG Banach space is Lin- delöf. Orihuela [323, Corollary 6] followed this method to prove also that a dual Banach space is weakly Lindelöf if and only if its weak∗ dual unit ball is a Corson compact space. Moreover, if E is a dual Banach space that is weakly Lindelöf, the product E × E is weakly Lindelöf [323, Theorem C]. We refer the reader to [13], [84], [86], [91] and [322] (and references therein) concerning weakly countably determined (WCD) Banach spaces and weakly Lindelöf determined (WLD) Banach spaces (providing larger classes of weakly Lindelöf spaces than the class of WCG Banach spaces). The following approach, using arguments from [321], was suggested to the authors by V. Montesinos; see also [149], [5].

Lemma 12.1 Let E be a Banach space. Let Δ ⊂ E be a set such that Δ is a linear subspace, and let ∇⊂E be a set that 1-norms Δ (i.e., for all x ∈ Δ, x= | | := { ∈∇; = } ⊕∇ supb ∈B∇ x,b , where B∇ b b 1 ). Then Δ ⊥ is a topological direct sum, and P =1, where P : Δ ⊕∇⊥ → Δ is the canonical projection.

~~Proof Obviously Δ ∩∇⊥ ={0},soΔ ⊕∇⊥ is an algebraic direct sum. Moreover, given x ∈ Δ and y ∈∇⊥, one has~~

~~P(x+ y)=x= sup |x,b | = sup |x + y,b | ≤ x + y, b ∈B∇ b ∈B∇ and hence P =1 and the direct sum is topological (in particular, closed).~~

w∗ Lemma 12.2 Let E be a Banach space, Δ, and ∇ be as in Lemma 12.1, with ∇ ∗ ∗ ⊥ w w a linear subspace. Then Δ ⊕∇⊥ = E if and only if Δ ∩ ∇ ={0}, where ∇ denotes the closure of ∇ in σ(E,E).

w∗ Proof The necessary condition is clear. Assume now that Δ⊥ ∩ ∇ ={0}.Letx ∈ ∗ ⊥ w E be such that x = 0. Then x ∈ Δ ∩∇ ,sox = 0. It follows that Δ⊕∇⊥ Δ⊕∇⊥ is dense in E; since, by Lemma 12.1, it is closed, we obtain the conclusion.

~~For a subset S ⊂ E of a Banach space E, denote n spanQ S := x : x = aizi,ai ∈ Q,zi ∈ S,1 ≤ i ≤ n, n ∈ N . i=1~~

~~We say that Y ⊂ E is Q-linear if spanQ Y = Y . 262 12 Weakly Realcompact Locally Convex Spaces~~

E E Lemma 12.3 Let E be a Banach space, Φ0 : E → 2 , and Ψ0 : E → 2 be two at most countably valued mappings. Suppose A0 ⊂ E, B0 ⊂ E and card (A0) ≤ Γ , card (B0) ≤ Γ for some inﬁnite cardinal Γ . Then there exist Q-linear sets A,B, A0 ⊂ A ⊂ E, B0 ⊂ B ⊂ E , such that card (A), card (B) ≤ Γ, and Φ0(B) ⊂ A, Ψ0(A) ⊂ B.

~~Proof We construct by induction two sequences of sets~~

~~A0 ⊂ A1 ⊂ A2 ···⊂E, B0 ⊂ B1 ⊂ B2 ···⊂E as follows. Having constructed A0 ...An, B0 ...Bn,weset~~

~~An+1 := spanQ(An ∪ Φ0(Bn)), Bn+1 := spanQ(Bn ∪ Ψ0(An)).~~

~~Finally, we set ∞ ∞ A := An,B:= Bn. n=0 n=0 That A and B satisfy the required properties is obvious.~~

Proposition 12.6 Let E be a WCG Banach space generated by an absolutely convex and weakly compact set K. Let A0, B0, Ψ0, Φ0 and Γ be as in Lemma 12.3. w∗ Then, there exist sets A and B such that card(A) ≤ Γ , card(B) ≤ Γ , A is linear, B is also linear, E = A ⊕ B⊥, the canonical projection P : E → A satisﬁes P =1 w∗ and P b = b for all b ∈ B .

~~Proof Given x ∈ E ,letϕ1(x ) be an element in K such that ϕ1(x ), x =sup |K,x |.~~

~~Put Φ(x ) = Φ0(x ) ∪{ϕ1(x )} for every x ∈ E .Givenx ∈ E,letψ1(x) ∈ BE such that |x,ψ1(x)| = x. Put~~

Ψ(x):= Ψ0(x) ∪{ψ1(x)} for every x ∈ E. We apply Lemma 12.3 to the sets A0 and B0 and Ψ and Φ.We obtain sets A and B as in the proof of Lemma 12.3. w First of all, from the construction in Lemma 12.3, it follows that A and B are linear sets. The set B 1-norms A. Indeed, let a ∈ A. There exists n ∈ N such that a ∈ An. Then ψ1(a) ∈ Bn satisﬁes |a,ψ1(a)| = a. This proves the assertion. By Lemma 12.1, we get that A ⊕ B⊥ is a direct sum. Let

~~w∗ x ∈ A⊥ ∩ B . 12.4 Every WCG Fréchet space is weakly K-analytic 263~~

~~w∗ μ(E,E) By the Mackey–Arens theorem, we have B = B , where μ(E,E) is the Mackey topology on E. Therefore, for any ε>0, there exists y ∈ B such that sup |K,x − y | <ε.~~

~~Then supK,x ≤supK,x − y +supK,y <ε+ supK,y = ε +ϕ1(y ), y =ε +ϕ1(y ), y − x ≤ε + supK,y − x < 2ε.~~

As ε>0 was arbitrary, we get x| ≡ 0, and so x = 0. This proves that ∗ K ⊥ w A ∩ B ={0}, and so, by Lemma 12.2,wehaveE = A ⊕ B⊥, and the canonical projection P : E → A satisﬁes P =1. Let

~~∗ w ⊥ b ∈ B (= (B⊥) ).~~

~~∗ w Put x = a + b, where a ∈ A and b ∈ B⊥. Then, for all b ∈ B ,wehave~~

~~x,Pb=Px,b=a,b=a + b,b=x,b, and hence P b = b.~~

~~Now we are ready to prove the following result due to Preiss and Talagrand.~~

~~Theorem 12.7 (Preiss–Talagrand) Every WCG Banach space is a weakly Lindelöf space.~~

Proof First observe that it is enough to choose countable A0 to get A countable, clearly dense in A and such that, for every element x ∈ Ψ0(A),wehaveP x = x . Let U := {Vt : t ∈ T } be an open cover of (E, σ (E, E )). For each x ∈ E,letpx be the supremum of all positive numbers p such that the open ball B(x,p) ⊂ Vt for −1 some t ∈ T . Choose Vx ∈ U such that B(x,px2 ) ⊂ Vx. Assume that Vx is deﬁned by a ﬁnite set Kx ⊂ E . Set Ψ0(x) := Kx . We apply Proposition 12.6, where Φ0 : E → 2E is a countably valued map. We get a countable dense set A in P(E)(= A) and such that, for every element x ∈ Ψ0(A),wehaveP x = x . We prove that the countable family {Vx : x ∈ A} is a covering of E. Fix arbitrary x ∈ E.LetB(P(x),p) ⊂ VP(x). Choose z ∈ A such that

~~− z ∈ B(P(x),10 1p).~~

~~−1 This implies that pz > 9(10) p. Hence~~

~~−1 B(z,2(5) p) ⊂ Vz.~~

~~This yields that P(x)⊂ Vz. Hence, for some ε>0, we have~~

~~|f(P(x)− z)| <ε 264 12 Weakly Realcompact Locally Convex Spaces for each f ∈ Ψ0(z). Then~~

~~|f(z− x)|=|P (f )(z − x)|=|f(P(x)− P(z))|=|f(P(x)− z)| <ε.~~

~~Hence x ∈ Vz.~~

Since every regular Lindelöf space is paracompact (Morita), Theorem 12.7 implies that the weak topology of a WCG Banach space is paracompact. We complete this section with the following theorem due to Khurana [242]extending Talagrand’s corresponding theorem [393] for WCG Banach spaces.

Theorem 12.8 (Khurana) Let E be a Fréchet space that admits an increasing sequence of σ(E,E)-compact sets whose union is dense in E. Then (E, σ (E, E)) is K-analytic. Moreover, E is a Borel subset of (E,σ(E,E)), where E is the bidual of E.

Proof Since for every metrizable lcs E the space (E, σ (E, E)) is angelic, to prove that (E, σ (E, E)) is a K-analytic space it is enough to show (by Corollary 3.6) that (E, σ (E, E)) has a compact resolution. In a natural way, we identify (E, σ (E, E)) with a subspace of RE endowed with the product topology. Therefore x = (g(x))g∈E for each x ∈ E. For each f ∈ E , E the map Pf : R → R deﬁned by

~~Pf ((αg)g∈E ) := αf satisﬁes~~

~~Pf (x) = Pf ((g(x))g∈E ) = f(x) for any x = (g(x))g∈E ∈ E, and therefore the restriction of Pf to E is f . Let (Vn)n be a basis of closed, absolutely convex neighborhoods of zero in E such that~~

(n + 1)Vn+1 ⊂ Vn ∈ N RE ∈ N ∈ 0 for each n .LetV n be the closure of Vn in . Then, given n, p , f Vn , and zn+p ∈ V n+p,wehave Pf (zn+p) ≤ sup Pf (x) : x ∈ V n+p = sup Pf (x) : x ∈ Vn+p .

~~Hence~~

Vn 1 P (z + ) ≤ sup |f(x)| : x ∈ ≤ (12.2) f n p n + p n + p ∈ 0 ∈ for each f Vn and zn+p V n+p.Let(An)n be an increasing sequence of weakly = compact, absolutely convex subsets of E such that n An H is dense in E. Since H is dense in E and Vn is a neighborhood of zero in E,wehave

~~E ⊂ H + Vn ⊂ H + V n 12.4 Every WCG Fréchet space is weakly K-analytic 265 for each n ∈ N.If x ∈ H + V n : n ∈ N , there exists a sequence~~

(x = hn + zn)n (12.3) ∈ ∈ ∈ N ∈ 0 ∈ N with hn H and zn V n.Fixann . Then, for each f Vn and p,q ,we note, by (12.2) and (12.3), that f(hn+p − hn+q ) = Pf (x − zn+p) − Pf (x − zn+q ) = Pf (zn+q ) − Pf (zn+p) , and therefore (12.2) yields −1 f(hn+p − hn+q ) ≤ Pf (zn+p) + Pf (zn+q ) ≤ 2n

~~∈ 0 for each f Vn . The uniformity implies that the sequence (hs)s is Cauchy in the Fréchet space E, and hence it has a limit h ∈ E. Then lims→∞ hs = h.From(12.2), it follows that~~

~~Pf (x) = lim Pf (hs + zs) = lim f(hs) + Pf (zs) = f(h)= Pf (h). s→∞ s→∞~~

~~Hence x = h in RE . Since x = h ∈ E, E = H + V n : n ∈ N , and then E = Am + Vn : m ∈ N : n ∈ N .~~

~~N Therefore, E admits a resolution {Kα : α ∈ N } with = + : ∈ N Kα Amn V n n~~

= ∈ NN RE + ∈ N for each α (mn) .In ,thesetsAmn V n for n are closed. E We claim that the closed set Kα is bounded in R . Indeed, if f ∈ E , there exists ∈ N ∈ 0 n such that f Vn and {| | : ∈ } ≤ : ∈ + sup f(x) x Kα sup Pf (x) x Amn V n ≤ | | : ∈ + : ∈ sup f(x) x Amn sup Pf (x) x V n = | | : ∈ + : ∈ ≤ + sup f(x) x Amn sup Pf (x) x Vn kmn 1, where

~~:= | | : ∈ kmn sup f(x) x Amn . 266 12 Weakly Realcompact Locally Convex Spaces~~

E This proves that the closed set Kα is compact in R , and it is also com- { } pact in (E, σ (E, E )). We proved that Kα : α ∈ N is a compact resolution in + RE + (E, σ (E, E )). Finally, since Amn V n is closed in ,theset(Amn V n) E is closed in (E,σ(E,E)), and so H + V n ∩ E is a Borel set in (E,σ(E,E)). Since E = H + V n ∩ E : n ∈ N , we deduce that E is a Borel set in (E,σ(E,E)).

~~Theorem 12.8 combined with Theorem 9.16 is applied to provide the following proposition.~~

~~Proposition 12.7 Let E be a WCG Baire lcs. Then E is a Fréchet space if and only if (E, σ (E, E)) is K-analytic.~~

Proof Assume E is a Fréchet space. By Theorem 12.8, the space E is weakly K- analytic. Conversely, assume (E, σ (E, E)) is K-analytic. Hence (E, σ (E, E)) admits a compact resolution. By Theorem 9.16, the space E is metrizable. Since E has a σ(E,E)-compact resolution and the original topology of E has a basis of neighborhoods consisting of σ(E,E)-closed sets, the space E admits a complete resolution. By Corollary 7.1, the space E is complete.

By Proposition 12.7, we deduce that the separable space X := RR is not K- analytic (since X is WCG Baire and nonmetrizable). Theorem 12.8 can also be deduced from Talagrand’s theorem [393] for WCG Banach spaces; see [79]. Indeed, let E be a WCG Fréchet space and (pn)n a fundamental sequence of continuous semi- ∈ N := −1 norms deﬁning the original topology of E. For each n , deﬁne En E/pn (0) + −1 := ∈ endowed with the normed topology x pn (0) pn(x) for each x E.LetFn be the completion of En for each n ∈ N. Clearly, E is linearly homeomorphic to a := closed vector subspace of the product F n Fn. Since each En is a continuous image of E, each En is a WCG space; hence Fn is a WCG Banach space. Now, by Talagrand’s theorem, each Fn is weakly K-analytic. Hence F is weakly K-analytic. Consequently, E is weakly K-analytic.

~~12.5 Amir–Lindenstrauss theorem~~

This short section, motivated by the previous one, provides a theorem of Amir and Lindenstrauss [270] stating that every nonseparable reﬂexive Banach space contains a complemented separable subspace. This result motivates us to study the property called the controlled separable projection property (CSPP). 12.5 Amir–Lindenstrauss theorem 267

~~The proof of Theorem 12.9 is due to Yost and uses some ideas from the proof of a more general result due to Valdivia [428, Lemma 1].~~

Theorem 12.9 (Amir–Lindenstrauss) Let E be a WCG Banach space and A0 and B0 two countable subsets of E and E , respectively. Then there exists a norm- one projection P : E → E with a separable range such that A0 ⊂ P(E) and B0 ⊂ P (E ). Consequently, for every separable subspace F in E there exists a closed, separable subspace G of E containing F and such that there is a norm-one projection P of E onto G.

~~Proof Let K be an absolutely convex, weakly compact subset of E whose linear span is dense in E. By the Hahn–Banach theorem, for each x ∈ E there exists a functional fx ∈ E such that~~

~~fx=1,fx(x) =x. Next, using the fact that K is σ(E,E )-compact, for every f ∈ E there exists xf ∈ K such that~~

~~|f(xf )|=max |f(x)|. x∈K By [C] we denote the Q-linear span of a set C (in E or E). Deﬁne~~

~~A1 := [A0 ∪{xf : f ∈ B0}],B1 := [B0 ∪{fx : x ∈ A1}], and next~~

A2 := [A1 ∪{xf : f ∈ B1}],B2 := [B1 ∪{fx : x ∈ A2}], and so on. Then, set ∞ ∞ A := An,B:= Bn. n=0 n=0 Clearly, A and B are countable Q-linear subspaces of E and E, respectively. Note also that for x ∈ A one has fx ∈ B, and for f ∈ B the corresponding xf belongs to A.Ifx ∈ A and y ∈ B⊥ ⊂ E,

~~x=fx(x) = fx(x + y) ≤x + y.~~

Then x≤x + y for each x ∈ A and y ∈ B⊥. This implies that A ∩ B⊥ ={0}, so the map P : A + B⊥ → A, x + y → x, is a surjective linear projection of norm one. Hence A + B⊥ is a closed linear subspace of E. To complete the proof, we need to show that A+B⊥ = E. It is enough to show that A + B⊥ is a (weakly) dense subspace of E. Choose arbitrary f ∈ E and assume that f |(A + B⊥) = 0. Then f |A = 0 and f |B⊥ = 0. Then f ∈ B⊥⊥ ⊂ E. 268 12 Weakly Realcompact Locally Convex Spaces

~~Since B⊥⊥ = B, where the closure is taken in the Mackey topology μ(E,E) of E,forε>0 there exists g ∈ B such that |f(x)− g(x)| <εfor each x ∈ K. Since xg ∈ A ∩ K and f |A = 0,~~

~~max |g(x)|=|g(xg)|=|f(xg) − g(xg)| <ε. x∈K~~

~~This implies that |g(x)| <εfor all x ∈ K. Consequently, |f(x)| < 2ε for all x ∈ K. Since the linear span of K is dense in σ(E,E), f = 0.~~

⊂ ⊂ A Banach space E such that for each sequence (xn)n E and (xn)n E there exists a continuous projection P : E → E with separable P(E) such that (xn)n and (xn)n are contained in P(E) and P (E ), respectively, will be said to have the controlled separable projection property (CSPP); see [417], [40]or[163]. By the results of Section 19.12, it will follow that every WLD Banach space satisﬁes the CSPP, and the Banach space C[0,ω1] provides a concrete example of a space where the CSPP is not the WLD (see Corollary 12.9). The last property follows from the fact that [0,ω1] is not Corson-compact. Recall here that a Banach space E is called weakly Lindelöf determined (WLD) if its dual unit ball is Corson compact when endowed with the weak∗ dual topology. This is a weaker property than being weakly countably determined WCD. Indeed, a Banach space E is WCD if E with the weak topology σ(E,E) is a Lindelöf Σ-space; see Proposition 3.5 and [15], [147]. Note that every WCG Banach space is WCD. Below we provide a couple of Banach spaces C(K) that do not satisfy the CSPP, [163]; hence they are not WLD. We shall say that a compact space K is countably measure determined if there exists a sequence (μn)n of Radon probabilities on K in C(K) such that ={ } n ker μn 0 . Clearly, every separable compact space enjoys this property. We note the following result from [163].

~~Proposition 12.8 Let K be a compact space, and assume that C(K) satisﬁes the CSPP. Then K does not contain a nonmetrizable countably measure determined closed subset.~~

Proof Assume that D is a closed nonmetrizable subset of K that is countably measure determined. Let (μn)n be a sequence of Radon probabilities on D with ={ } n ker μn 0 .Let(λn)n be a sequence of Radon measures on K such that each λn extends μn and supp μn = supp λn for each n ∈ N. By the CSPP, there exists a continuous projection P : C(K) → C(K) such that P(C(K)) is separable ⊥ and P (C(K) )(= (ker P) ) contains the sequence (λn)n. Deﬁne the linear surjective map T : P(C(K))→ C(D) by the formula T(P(f)):= f |D, f ∈ C(K). Note that the map T is well deﬁned. Indeed, if f,g ∈ C(K) with P(f)= P(g), f − g ∈ ker P. This yields that

(μn,(f − g)|D) = (f − g)dμn = (f − g)dλn = (λn,(f − g)) = 0 D K 12.5 Amir–Lindenstrauss theorem 269 for each n ∈ N. Hence (f − g)|D = 0, so T is well deﬁned. Note that T is continuous. Indeed, if f ∈ Cc(K),wehavef − P(f)∈ ker P and then f |D = (P (f ))|D. Hence T(P(f))=f |D=(P (f ))|D≤P(f). This shows that the separable quotient P (C(K))/ ker T is isomorphic to a nonseparable space C(D), yielding a contradiction.

~~Corollary 12.6 If K is a compact space that contains a nonmetrizable separable subset, the space C(K) does not have the CSPP.~~

~~Corollary 12.7 If a compact space K contains βN, the space C(K) does not have the CSPP.~~

~~Since every metric compact scattered space is countable, we have the following corollary.~~

~~Corollary 12.8 Let K be a scattered compact space that contains a separable uncountable subset. Then C(K) does not have the CSPP.~~

Summarizing, we note the following: (A) Every WLD Banach space satisﬁes the CSPP. Therefore, the space C(K) in Corollary 12.8 is not WLD. (B) The Banach space C([0,ω1]) has the CSPP and is not WLD (see Corol- lary 12.9). Let K be a compact scattered space, and let μ be a nonnegative Radon measure on K.LetA ⊂ K be a nonzero Borel subset of K. Since μ is regular, there exists a compact set B ⊂ A with μ(B) > 0. Let λ := μ|B. Then there exist an open set U ⊂ K and t ∈ supp λ such that

~~U ∩ supp λ ={t}.~~

~~This implies that λ(U ∩ B) > 0. Since~~

U ∩ B ={t}∪(U ∩ B \ (supp λ)) , one has μ({t}) = λ({t}) = λ(U ∩ B). This shows that {t} is an atom in μ. Hence μ has a countable set of atoms. Let L be a locally compact scattered Hausdorff space, and let K := L ∪{∞}be the one-point compactiﬁcation of L. We have the following proposition.

Proposition 12.9 [163] Assume that K := L ∪{∞}satisﬁes the condition (x) (i.e., the closure of every countable subset of L is countable). Then, for every Radon measure μ on K, there exists a countable subset N ⊂ L such that supp μ ⊂ N ∪{∞}. 270 12 Weakly Realcompact Locally Convex Spaces

~~We also need the following two additional easy facts; see [163].~~

~~Lemma 12.4 Let G be an almost-clopen subset of K := L ∪{∞}(i.e., an open set such that G ⊂ G ∪{∞}). Then, for each f ∈ C(K), the function f · 1G + f(∞) · 1K\G ∈ C(K).~~

Lemma 12.5 If K := L ∪{∞}satisﬁes the property (xx) (i.e., every closed subset of K contained in L is countable), for each f ∈ C(K) there exists a countable set N ⊂ L such that f(K\ N)={f(∞)}.

~~We are ready to prove the following general result [163].~~

Proposition 12.10 If K := L ∪{∞}satisﬁes the conditions (xx) and (xxx) (i.e., for each countable subset N ⊂ L there exists a countable almost-clopen set G with N ⊂ G), the space C(K) satisﬁes the CSPP.

Proof We may assume that L is uncountable; otherwise K would be countable and metrizable (every compact countable space is metrizable!), so C(K) would be separable, thus satisfying the CSPP. Let (fn)n and (μn)n be sequences in C(K) and C(K) , respectively. By Lemma 12.5 and Proposition 12.9, there exists a countable set N ⊂ L such that

fn(K \ N)={fn(∞)}, supp μn ⊂ N ∪{∞}, for each n ∈ N. By the condition (xxx), there is a countable almost-open set G containing N. We deﬁne a linear and continuous map P : C(K) → C(K) by the formula P(f):= f · 1G + f(∞) · 1K\G for each f ∈ C(K). Note also that

~~2 P (f ) = P(f)· 1G + P(f)(∞) · 1K\G =~~

(f · 1G + f(∞) · 1K\G) · 1G + f(∞) · 1K\G = P(f). Case 1. G = G ∪{∞}. Then P(C(K))is linearly homeomorphic to C(G). Since G is metrizable, C(G) is separable, and hence P(C(K))is separable, too. Case 2. G = G. In that case, C(G) is linearly homeomorphic to a closed, separable hyperplane of P(C(K)). Hence P(C(K))is also separable. Finally, note that (fn)n ⊂ P(C(K))and ⊥ (μn)n ⊂ P (C(K) ) = (ker P) for each n ∈ N. The ﬁrst claim is clear. If f ∈ ker P , then

~~μn(f ) = fdμn = fdμn = 0 supp μn G∪{∞} since f |G = 0 and f(∞) = 0. 12.6 An example of Pol 271~~

Now we are ready to show that the CSPP does not imply the Lindelöf property for the weak topology of a Banach space. For a more general statement concerning the CSPP and compact lines, see Theorem 18.3 below.

~~Corollary 12.9 The space C([0,ω1]) satisﬁes the CSPP and is not WLD.~~

Proof The space C([0,ω1]) is not weakly Lindelöf. Indeed, assume that C([0,ω1]) is weakly Lindelöf. Then, by Theorem 9.11, the space [0,ω1] has countable tightness. This is impossible since the closure of the set A := [0,ω1]\{ω1} contains ω1 and the closure of any countable subset of A does not contain ω1. This provides a contradiction. Clearly, [0,ω1] is the one-point compactification of the locally compact space [0,ω1). It is enough to prove that [0,ω1] satisfies the conditions (xx) and (xxx), and then apply Proposition 12.10. [0,ω1] satisfies the condition (xx). Indeed, let A ⊂[0,ω1) be closed in [0,ω1]. Since ω1 ∈ A, there exists α<ω1 such that A ∩[α, ω1]=∅. This shows that A is countable since A ⊂[0,α]. [0,ω1] satisfies the condition (xxx). Indeed, let N be a countable subset of [0,ω1). Choose α<ω1 such that N ⊂[0,α]. Then G := [0,α] is a clopen countable set containing N.

~~12.6 An example of Pol~~

~~The previous results may suggest the following natural question.~~

~~Problem 12.2 Is a weakly Lindelöf Banach space a WCG Banach space?~~

This problem has been announced by Corson [103]; see also [271]. Talagrand [388] gave an example of a compact space X such that C(X) is weakly K-analytic and is not a WCG Banach space. Another example of this type was obtained by Pol. Pol [336] showed that there exists a Banach space C(X) over a compact scattered space X such that C(X) is weakly Lindelöf and is not a WCG Banach space (see Theorem 12.10). This example answers also (in the negative) some questions of Corson [103] posed by Benyamini, Rudin and Wage in [49, Problem 7]. Theorem 12.10 will be used (in the next section) to show that (gDF)-spaces are not in the class G. Results in this section are due to Pol [336]. A space E is said to have the strong condensation property if for each uncountable subset A of E there exists an uncountable subset C ⊂ A that is concentrated around a point c ∈ E (i.e., the set C\V is at most countable whenever V is a neighborhood of c). Let Ω =[1,ω1[ be the set of all countable ordinals endowed with the usual order. Then, for each α ∈ Ω, the interval [1,α]:={β ∈ Ω : β ≤ α} is countable. Therefore, if D is an uncountable subset of [1,ω1[, there exists an injective map ϕ :[1,ω1[→ D 272 12 Weakly Realcompact Locally Convex Spaces such that β<ϕ(β)<ϕ(α)for each countable ordinal α, β such that β<α;themap ϕ can be determined by transﬁnite induction, where ϕ(α) is the ﬁrst ordinal of the set {γ ∈[δ,ω1[: xγ ∈ D}, and δ := sup{α, sup{ϕ(γ) : γ<α}}. Note the following proposition.

Proposition 12.11 A topological space X has the strong condensation property if and only if each net {xα : α ∈[1,ω1[} in X admits a convergent subnet {xϕ(α) : α ∈[1,ω1[} directed by Ω. Additionally, we may assume that the map ϕ :[1,ω1[→ [1,ω1[ satisﬁes α<ϕ(α)<ϕ(α) if α<α.

Proof Using transﬁnite induction, we note that an uncountable set A contains an uncountable subset A = {xα : α ∈[1,ω1[}, where xα ∈ A\∪{xβ : β ∈[1,α[}. Then, if C = {xα : α ∈ D} is an uncountable subset of A, we have that C contains an uncountable set C = {xϕ(α) : α ∈[1,ω1[}, where ϕ :[1,ω1[→ D is the map as above. {xϕ(α) : α ∈[1,ω1[}is a subnet of {xα : α ∈[1,ω1[}, and C is concentrated around a point c ∈ E if and only if = limα∈[1,ω1[ xϕ(α) c. Therefore, we proved that a net {xα : α ∈[1,ω1[}in X with the strong condensation property has a convergent subset {xϕ(α) : α ∈[1,ω1[}if the set {xα : α ∈[1,ω1[} is uncountable. If {xα : α ∈[1,ω1[} is countable, there exists an uncountable subset D in [1,ω1[ such that xα = xβ , where α, β ∈ D, and the map ϕ :[1,ω1[→ D is as above. This proves that the constant net {xϕ(α) : α ∈[1,ω1[}is a convergent subnet of the net {xα : α ∈[1,ω1[}. The rest of the proof is clear.

If a topological space X has the strong condensation property and the weight w(X) of X is at most ℵ1, the space X is Lindelöf. This follows from the fact that if {Bα : α ∈[1,ω1[}is a basis of X and X is not Lindelöf, there exists a net {xα : α ∈ [1,ω1[}in X with

xα ∈ Y \{Bβ : β ≤ α} for each α ∈[1,ω1[ (since X is not covered by {Bβ : β ≤ α} for each α ∈[1,ω1[). As each Bα only contains a countable number of points of {xα : α ∈[1,ω1[}, the net {xα : α ∈[1,ω1[}does not admit a convergent subnet. Then X does not have the strong condensation property. 12.6 An example of Pol 273

~~Let Γ ⊂ Ω be the set of all nonlimit ordinals, and let Λ = Ω\Γ . Attach to each γ ∈ Λ an increasing sequence (sλ(n))n in Γ such that~~

~~lim sλ(n) = λ. n→∞ Endow the set Ω with the following topology: The points from Γ are isolated, and the basic neighborhoods of a point γ ∈ Λ are of the form~~

~~Wγ (n) ={γ }∪{sγ (m) : m n}.~~

Therefore, if K is a compact subset of Ω,thesetK ∩ Λ is finite because for each γ ∈ K ∩ Λ the set Wγ (n) ∩ Λ is unitary and the points of Γ are isolated. Let X := Ω ∪{ω1} be the one-point compactification of the locally compact space Ω, where now the first uncountable ordinal ω1 is the point at infinity.The compact space X is scattered. Our goal is to prove that the Banach space C(X) is weakly Lindelöf. Recall that for a compact scattered space X the weak topology of C(X) coincides on the unit ball with the topology of Cp(X). Therefore, the space C(X) is weakly Lindelöf if and only if Cp(X) is Lindelöf. As usual, |A| denotes the cardinal of the set A and 1X is the constant function on X with 1X(X) ={1}.To avoid misunderstanding, the space Cp(X) will sometimes be denoted by Cp(X, R). Let D ={0, 1} be endowed with the discrete topology. Then Cp(X, D) is the subset of Cp(X, R) determined by the functions with the range in D. We need the following lemma.

Lemma 12.6 Assume that s(α) ∈ Γ for each α ∈[1,ω1[ and that χs(α) is the characteristic function of {s(α)}. Then the net {χs(α) : α ∈[1,ω1[} admits a convergent subnet {χs(ϕ(α)) : α ∈[1,ω1[} in Cp(X, D).

Proof Indeed, by Proposition 12.11, we note that if D := {s(α) : α ∈[1,ω1[} is uncountable, there exists an injective map ϕ :[1,ω1[→ D such that α<ϕ(α)< ϕ(β) if α<β.Then, the limit of the net {χs(ϕ(α)) : α ∈[1,ω1[}in Cp(X, D) is the null function. From Proposition 12.11, it follows that if {s(α) : α ∈[1,ω1[} is countable, there exists an uncountable subset D in [1,ω1[ such that s(α) = s(β) if α, β ∈ D. Therefore there exists a constant map ϕ :[1,ω1[→ D with ϕ(α) = ζ for each α ∈[1,ω1[. Then, the limit of the net

~~{χs(ϕ(α)) : α ∈[1,ω1[} in Cp(X, D) is the characteristic function χζ .~~

~~This yields the following corollary. 274 12 Weakly Realcompact Locally Convex Spaces~~

~~−1 Corollary 12.10 Let G0 := {f ∈ Cp(X, D) : f (1) ⊂ Γ } be endowed with the induced topology of Cp(X, D). Then G0 has the strong condensation property.~~

Proof We prove that the net {fα : α ∈[1,ω1[}in G0 contains a subnet {fα : α ∈ [ [} −1 1,ω1 that converges in G0. By the compactness of fα (1) (in the discrete sub- −1 ∈ N space Γ ), we deduce that the compact set fα (1) is ﬁnite. There exists m such { ∈[ [: −1 = } that the set α 1,ω1 fα (1) m is uncountable. Hence, taking a subnet if necessary, we may assume that −1 ={ ··· } fα (1) s1(α), s2(α), ,sm(α) for each α ∈[1,ω1[. If m = 1, the existence of the convergent subnet follows from Lemma 12.6.As- sume, as the induction hypothesis, that a net {gα : α ∈[1,ω1[}in G0, such that | −1 |= − ∈[ [ { : ∈[ [} gα (1) m 1 for each α 1,ω1 , has a convergent subnet gϕ(α) α 1,ω1 in G0. Then, if {fα : α ∈[1,ω1[}is a net in G0 such that −1 ={ ··· } fα (1) s1(α), s2(α), ,sm(α) ∈[ [ { : ∈[ [} for each α 1,ω1 , then, by Lemma 12.6, the net χs1(α) α 1,ω1 admits a { : ∈[ [} convergent subnet χs1(ϕ(α)) α 1,ω1 in G0. By the induction hypothesis, the net { − : ∈[ [} fϕ(α) χs1(ϕ(α)) α 1,ω1 has a convergent subnet { − : ∈[ [} fϕ(ψ(α)) χs1(ϕ(ψ(α))) α 1,ω1 in G0. Therefore, the net {fϕ(ψ(α)) : α ∈[1,ω1[}also converges in G0.

~~Now we prove the following proposition.~~

~~Proposition 12.12 Let D ={0, 1} be endowed with the discrete topology. Then the space Cp(X, D) has the strong condensation property.~~

~~Proof For each n ∈ N,set −1 Gn := {f ∈ Cp(X, D) : f (1) ∩ Λ = n}.~~

~~By the continuity of the map ϕ : Cp(X, D) → Cp(X, D) deﬁned by ϕ(f) := 1X − f , from the equality~~

Cp(X, D) ={f ∈ Cp(X, D) : f(ω1) = 0}⊕{f ∈ Cp(X, D) : f(ω1) = 1}, and since {f ∈ Cp(X, D) : f(ω1) = 0}= Gn, n 12.6 An example of Pol 275 we note that Cp(X, D) has the strong condensation property, provided that for each n ∈ N the following holds: (*) Every net {fα : α ∈[1,ω1[}in Gn has a subnet {fϕ(α) : α ∈[1,ω1[}that converges in Cp(X, D). In Corollary 12.10, we proved (*) for n = 0. We prove the condition (*) by induction. Assume (*) holds in Gn, and assume that {fα : α ∈[1,ω1[}is a net in Gn+1. Then −1 ∩ ={ ··· } fα (1) Λ s1(α), s2(α), ,sn+1(α) for each α ∈[1,ω1[. Taking a subnet, if necessary, we may assume that s1(α) = s1 for each α ∈[1,ω1[,or

~~sup{s1(β) : β<α}~~

~~{f − χ : α ∈[1,ω [}. ϕ(α) s1 1~~

~~This implies that {fϕ(α) : α ∈[1,ω1[}converges. For the other case (i.e., if~~

~~δ(α) := sup{s1(β) : β<α}~~

~~This yields the following corollary.~~

~~N Corollary 12.11 Cp(X, D ) is a Lindelöf space.~~

Proof Since ω(X) ≤ℵ, we deduce that ω(C (X, D)) ≤ ω(X2) ≤ℵ2 =ℵ .This 1 p 1 1 N and Proposition 12.12 imply that Cp(X, D) has the strong condensation property, N and its weight is at most ℵ . Therefore Cp(X, D) is a Lindelöf space. Now we get 1 N N the following corollary from the fact that Cp(X, D) and Cp(X, D ) are homeomorphic.

~~Corollary 12.12 Let Q be a countable subset of DN and P = DN\Q. Then Cp(X, P ) is a Lindelöf space. 276 12 Weakly Realcompact Locally Convex Spaces~~

Proof Set Q ={q1,q2, ···}, and for each i ∈ N let (Vn(qi))n be a base of open N neighborhoods of qi .AsCp(X, D ) has the strong condensation property, we deduce that for each n ∈ N the closed subspace

N N Ei,n := {f ∈ Cp(X, D ) : f(X)∩[D \Vn(qi)]=∅} has the strong condensation property. Then the union N Ei := Ei,n ={f ∈ Cp(X, D ) : f(X)∩{qi}=∅} n also has the strong condensation property. Also, the product i∈N Ei and its diagonal have the strong condensation property. This and the fact that the space Cp(X, P ) ℵ has weight at most 1 completes the proof.

It is known (see [336, p. 281, proof of Lemma 1] and [262, 40, VII]) that, if N a countable set Q is dense in D , the space Cp(X, R) is a continuous image of Cp(X, P ). On the other hand, for a compact space S, the Banach space C(S) is WCG if and only if S is Eberlein compact. Also, Wage proved [410, Example p. 20] that the compact space X = Ω ∪{ω1} is not Eberlein compact. Hence the Banach space C(X) is not WCG. Therefore, we have proved the following theorem.

~~Theorem 12.10 (Pol) The space C(X,R) is Lindelöf and has the strong condensation property, and C(X,R) is not a WCG Banach space.~~

~~12.7 More about Banach spaces C(X) over compact scattered X~~

~~In this section, we use the previous example of Pol to show that in general (gDF )- spaces do not belong to the class G. First we prove the following proposition.~~

~~Proposition 12.13 Let X be a scattered compact space such that C(X) is weakly Lindelöf. Then the weak∗ dual of C(X) has countable tightness.~~

Proof Let τp and τσ := σ(C(X),C(X)) be the original topology of Cp(X) and the weak topology of C(X), respectively. Since (C(X), τσ ) is Lindelöf, Cp(X) is also Lindelöf. Let B be the closed unit ball in C(X). Since X is scattered, we have τp|B = τσ |B by [372, Corollary 19.7.7]. As a direct proof, clearly τp|B ≤ τσ |B. The argument used in the proof of The- orem 9.2 (i) is applied to show that every sequence in B that converges in τp also converges in τσ . Hence the identity map from (B, τp) onto (B, τσ ) is sequentially continuous. On the other hand, since X is scattered, the space Cp(X) is Fréchet– Urysohn (see Proposition 14.1). As (B, τp) is Fréchet–Urysohn, we deduce easily 12.7 More about Banach spaces C(X) over compact scattered X 277 that the (sequentially continuous) identity map (B, τp) → (B, τσ ) is continuous. We | = | n| n = n| n ∈ N n := proved τp B τσ B. Then τp B τσ B for each n , where B 1≤i≤n B n n n := and τp , τσ denote the own product topologies on C(X) 1≤i≤n C(X). Since X is compact and scattered, X is zero-dimensional, and then [27, Theorem IV.8.6] n n ∈ N is applied to deduce that (Cp(X) ,τp ) is a Lindelöf space for each n .The n n n space B (being closed in τp ) is also a Lindelöf space. Hence B is Lindelöf in n n ∈ N n = n n (C(X) ,τσ ) for each n .WehaveC(X) m mB , and each mB is a Lin- n n n ∈ N delöf space in τσ , so the space (C(X) ,τσ ) is a Lindelöf space for each n . Proposition 9.9 implies that Cp((C(X), τσ )) has countable tightness. Hence the space C(X) , σ (C(X) , C(X)) ⊂ Cp(C(X), τσ ) has countable tightness.

~~We also need the following result due to R. Pol [336].~~

~~Proposition 12.14 Every weakly K-analytic Banach space C(X) over a compact scattered space X is a WCG Banach space.~~

Let X := Ω ∪{ω1} be the compact scattered space (the one-point compactiﬁcation of the locally compact space Ω) considered in Section 12.6. By Theorem 12.10 and proof of Proposition 12.13, the Banach space C(X) is weakly Lindelöf and not WCG by [410]. Since every weakly K-analytic Banach space C(X) over a compact scattered X is a WCG Banach space (Proposition 12.14), the space C(X) is not weakly K-analytic. Hence we have the following (see Proposition 12.13).

~~Proposition 12.15 C(X) is weakly Lindelöf and is not weakly K-analytic. The weak∗ dual of C(X) has countable tightness.~~

~~On the other hand, combining Proposition 12.13 with Example 9.4 (and its proof), we note the following example.~~

Example 12.3 Assume the CH. There exists an uncountable compact scattered space X such that C(X) is weakly Lindelöf and C(X) is not a weakly Lindelöf Σ-space. Nevertheless, the weak∗ dual of C(X) has countable tightness.

Note also that examples of compact X such that C(X) are weakly Lindelöf and not weakly Lindelöf Σ (in other words, Corson compact, not Gul’ko compact) can be found in [15]; see also [13, Theorem 3.3] and Section 19.17 below. The preceding examples will be used to show that the (gDF)-spaces need not be in the class G. Following Ruess, an lcs E is a (gDF)-space if it has a fundamental sequence (Bn)n of bounded sets and is C-quasibarrelled (i.e., for every sequence (Un)n of absolutely convex closed neighborhood of zero such that for any bounded set B in ∈ N ⊂ ≥ = E, there exists p such that B Un for each n p,thesetU n Un is a neighborhood of zero). It is known that an lcs (E, ξ) is a (gDF )-space if and only if it has a fundamental sequence (Bn)n of bounded sets and ξ is the ﬁnest locally convex topology on E that agrees with ξ|Bn on each set Bp;see[328]. 278 12 Weakly Realcompact Locally Convex Spaces

Clearly, every (DF )-space is a (gDF )-space. Every (DF )-space is a dual metric space. Corollary 2.10 showed that Cc(X) is a (df )-space if and only if it has a ∞ fundamental sequence of bounded sets and Cc(X) is -barrelled. Hence a (df )- space Cc(X) is dual metric, so it belongs to the class G (we already showed that all dual metric spaces are in the class G). We prove that there exist (gDF)-spaces that are not in the class G. We need the following observation (see [350], [351]; for a simple proof, we refer to [328, Proposition 8.3.10]).

~~Proposition 12.16 Let E be a Fréchet space. Then (E ,τpc(E ,E)) is a (gDF)- space.~~

~~We complete this section with the following promised proposition.~~

~~Proposition 12.17 There exists a (gDF)-space that does not belong to the class G.~~

Proof Fix E := C(X), where X := Ω ∪{ω1} is the compact scattered space considered above. Assume (E ,τpc(E ,E))belongs to the class G. By Proposition 12.13, we know that (E,σ(E,E)),theweak∗ dual of C(X), has countable tightness. Then, by Theorem 12.2, the weak topology of C(X) is K-analytic, a contradiction with Proposition 12.15.

~~Proposition 12.17 also provides an example of a (gDF )-space that is not a dual metric space. Chapter 13 Corson’s Property (C) and Tightness~~

Abstract In this chapter, the class of Banach spaces having the property (C) (isolated by Corson) is studied. This property provides a large subclass of Banach spaces E whose weak topology need not be Lindelöf. We collect some results of Corson, Pol, Frankiewicz, Plebanek and Ryll-Nardzewski.

~~13.1 Property (C) and weakly Lindelöf Banach spaces~~

Let X be a completely regular Hausdorff space. We know already that the space n Cp(X) has countable tightness if and only if each ﬁnite product X is a Lindelöf space; see Theorem 9.9. This result leads to the following problem.

~~Problem 13.1 Assume that X is a Lindelöf space. Is it true that every compact subset of Cp(X) has countable tightness?~~

This interesting and difﬁcult question has been answered by Arkhangel’skii [27, Theorem IV.11.14] assuming the Proper Forcing Axiom. Problem 13.1 may also suggest another axiom formulated for Banach spaces.

~~Problem 13.2 Let E be a Banach space such that (E, σ (E, E)) is a Lindelöf space. Is the unit ball in E of countable tightness in σ(E,E)?~~

Problem 13.2 is strictly connected with Corson’s property (C);see[103]. A convex closed subset M of a Banach space E is said to have the property (C) if M veriﬁes one of the following equivalent conditions: (i) Every family F of complements of closed convex sets in M that covers M has a countable subfamily covering M. (ii) For every family A of closed convex subsets of M with the empty intersection, there is a countable subfamily B of A with the empty intersection. (iii) Every family A of nonempty closed convex subsets of M closed under countable intersections veriﬁes A = ∅. We shall say that a Banach space E has the property (C) if the above holds for E = M.If(E, σ (E, E)) is a Lindelöf space, the space E has the property (C) (since closed convex sets in E and σ(E,E) are the same). Hence every WCG

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 279 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_13, © Springer Science+Business Media, LLC 2011 280 13 Corson’s Property (C) and Tightness

Banach space has the property (C). Every nonseparable WCG Banach space has the property (C) and is not Lindelöf. We show, using Corson [103, Example 2] and Pol [334, Proposition 1], that there exist Banach spaces with the property (C) that are not weakly Lindelöf. See also [103, Examples 2, 3, 4]. Next, we show that the property (C) is a three-space property. We need the following results from [334].

Proposition 13.1 (Pol) The following conditions are equivalent for a Banach space E: (i) E has the property (C). (ii) If a family K of nonempty closed convex sets in E is closed under countable intersections, then for every σ>0 there is a ∈ E with dist (a, K) < σ for every K ∈ K . (iii) If a family K of nonempty closed convex sets in E is closed under countable intersections, then for every σ>0 there is a closed convex subset M with the property (C) such that dist (M, K) < σ for every K ∈ K . (iv) If a family K of nonempty closed convex sets in the closed unit ball BE in E is closed under countable intersections, then for every σ>0 there is a closed convex subset M with the property (C) such that dist (M, K) < σ for every K ∈ K . Proof (i) ⇒ (ii): If E has the property (C),wehave A = ∅. (ii) ⇒ (i): We need to prove that if C is a family of nonempty closed convex subsets of X closed under countable intersections, one has C = ∅. From (ii) we −1 deduce that there exists a1 ∈ E with dist (a1,C)<2 for every C ∈ C . Moreover, −i if for 1 ≤ i ≤ k we deﬁne ai ∈ E with dist (ai,C)<2 for every C ∈ C and −i+2 dist (ai−1,ai)<2 for 2 ≤ i ≤ k, (ii) applied to the family −k C ∩ ak + 2 BE : C ∈ C yields an element ak+1 ∈ E with −k −k−1 dist (ak+1,C∩ (ak + 2 BE)) < 2 for every C ∈ C . It is clear that −k−1 −k+1 dist (ak+1,C)<2 , dist (ak+1,ak)<2 .

Continuing this inductive procedure, we determine a Cauchy sequence (ak)k with limit a ∈ C for every C ∈ C. This completes the proof since a ∈ C . (ii) ⇒ (iii) is clear. (iii) ⇒ (ii): If (ii) fails, there exist a family K of nonempty closed convex subsets of E closed under countable intersections and a positive number ε such that for each a ∈ E there exists Ka ∈ K with dist (a, Ka) ≥ ε. By (iii) there exists a closed convex subset M of E with the property (C) such that dist (M, K) < ε2−1 for each K ∈ K .AsK is closed under countable intersections and M has the property (C), we deduce that the intersection of the family

~~−1 {(K + ε2 BE) ∩ M : K ∈ K } 13.1 Property (C) and weakly Lindelöf Banach spaces 281 is nonempty. Then there exists~~

−1 y ∈ K + ε2 BE for each K ∈ K . This implies that dist (y, K) < ε for each K ∈ K . Hence dist (y, Ky)<ε, a contradiction. (iii) ⇒ (iv) is clear. (iv) ⇒ (iii): Assume (iii) fails. Then there exist a family K of nonempty closed convex subsets of E closed under countable intersections and a positive number ε such that for each closed convex subset M of E with the property (C) there exists KM ∈ K with dist (M, KM ) ≥ ε. Since K is closed under countable intersections, there is a natural number n ∈ N such that K ∩ nBE = ∅ for every K ∈ K . Since dist (M, KM ∩ nBE) ≥ ε, then −1 −1 −1 dist n M,n [KM ∩ nBE] ≥ εn .

~~−1 −1 Therefore (iv) fails for the family {n [K ∩ nBE]:K ∈ K } and σ = εn .~~

~~Now we are ready to prove the following theorem.~~

~~Theorem 13.1 (Pol) Let F be a closed subspace of a Banach space E. If the spaces F and E/F have the property (C), the space E has the property (C).~~

Proof Assume F and E/F have the property (C).Letq : E → E/F be the quotient map. If E does not have the property (C), by Proposition 13.1 there exist ε>0, a family K of nonempty closed convex subsets of E that are closed under countable intersections, such that for every closed convex subset M with the property (C) there exists KM ∈ K with dist (M, KM ) ≥ ε. For each z ∈ E,thesetz + F is closed, convex and has the property (C) (by the assumption, F has the property (C)). Then there exists Kz+F ∈ K with

~~dist (z + F,Kz+F ) ≥ ε.~~

~~This implies that q(z)∈ / q(Kz+F ). Hence q(K) : K ∈ K =∅, contradicting that E/F has the property (C).~~

Theorem 8.1 and Theorem 13.1 provide a Banach space with the property (C) such that (E, σ (E, E)) is not a Lindelöf space by Lemma 6.1. Next, Corollary 13.1 will be used to characterize the property (C).

Corollary 13.1 Assume that a Banach space E does not have the property (C). Then there exists a subset A of the unit ball BE of E , >0, such that for every linear subspace M ⊂ E with the property (C) there exists f ∈ A vanishing on M, and for each countable C ⊂ A there is x ∈ E with x ≤1 and f(x)≥ for every f ∈ C. 282 13 Corson’s Property (C) and Tightness

Proof Let F be the family of all subspaces of E with the property (C). Since E does not have the property (C), by Proposition 13.1 (iii) there exists a family K of nonempty closed convex sets in E closed under countable intersections, >0, such that for each F ∈ F there exists KF ∈ K with dist (F, KF ) ≥ . Let B be the open unit ball in E. Since (F + B)∩ KF =∅, by the Hahn–Banach theorem, there exists a continuous linear functional fF ∈ E such that fF =1 and

sup fF (x) ≤ inf fF (y). ∈ x∈F +B y KF Then fF (x) = 0 for all x ∈ F and fF (y) ≥ for every y ∈ KF . Set A := {fF : ∈ } ={ } F F , and let C fF1 ,fF2 ,..., be a countable subset of A . Then the desired conditions are satisﬁed for >0 chosen above and x equal to a point of the nonvoid { : ∈ N} set KFn n .

We know (see Theorem 12.2) that if E is a metrizable lcs, its weak∗ dual (E,σ(E,E)) is a Lindelöf space. We characterize Banach spaces with the property (C) in terms of some tightness-type property for the weak∗ dual (E,σ(E,E)). Condition (ii) below can be formulated for the closed unit ball in E (instead of taking the whole dual space E).

Theorem 13.2 (Pol) For a Banach space E, the following assertions are equivalent: (i) E has the property (C). (ii) For each set A ⊂ E and each f ∈ A (the closure in σ(E,E)), there exists countable B ⊂ A such that f ∈ conv(B).

~~Proof (i) ⇒(ii): Fix f ∈ A. Set~~

Cg := {x ∈ E : g(x) ≥ f(x)+ 1} ∈ =∅ for each g A. Clearly, the sets Cg are closed and convex, and g∈ACg . Since ⊂ =∅ E has the property (C), there exists a countable subset B A with g∈B Cg . Then f ∈ conv(B). Indeed, otherwise, by the Hahn–Banach theorem, there exists ∈ ≥ + ∈ ∈ x E such that h(x) f(x) 1 for each h conv(B). Hence x g∈B Cg, a contradiction. (ii) ⇒(i): Assume E does not have the property (C). Note that 0 ∈ A, where A is deﬁned in the proof of Corollary 13.1. By this corollary, we deduce that 0 does not belong to conv(C) for each countable C ⊂ A. This yields a contradiction with the assumption (ii).

Combining Proposition 12.1 with Theorem 13.2, we deduce that, for a Banach space with the property (C), the weak topology σ(E,E) is realcompact. This fact has already been observed by Corson [103, Lemma 9]. Also, the following fact is clear; see [334, Theorem 5.1].

~~Proposition 13.2 If K is a compact space and the Banach space C(K) has the property (C), the space K has countable tightness. 13.1 Property (C) and weakly Lindelöf Banach spaces 283~~

Proof Fix A ⊂ K and x ∈ A. Set CB := {g ∈ C(K) : g|B = 0,g(x) = 1} for ω ω each B ∈ A , where A denotes the family of all countable subsets of A. Since =∅ B∈Aω CB and each set CBis closed and convex, the property (C)provides a ω =∅ ∈ := sequence (Bn)n in A such that n CBn . Then x B, where B n Bn.

~~We complete this section by showing that for Banach spaces E with the property (C) we have ck(H ) = k(H ) for every bounded set H ⊂ E;see[7, Proposition 2.6]. We need the following lemma from [7].~~

~~Lemma 13.1 Let x∗∗ ∈ E \ E and b ∈ R with d(x∗∗,E) > b > 0. Then 0 ∈ ∗ ∗∗ ∗ {x ∈ BE : x (x )>b}, where the closure is taken in σ(E ,E).~~

∗∗ Proof Let >0 and x1,x2,...,xn ∈ E. We may assume that b + b}. By the Hahn–Banach theorem, there exists ξ ∈ E with ξ(x) = 0 for each x ∈ E and ξ =1 with ξ(x∗∗) = d(x∗∗,E). By the classical Goldstein theorem (see ∗ [213]), there exists x ∈ BE (⊂ BE ) such that ∗ ∗ |ξ(xi) − x (xi)|=|x (xi)| < for 1 ≤ i ≤ n and |ξ(x∗∗) − x∗∗(x∗)| <.Then x∗ and −x∗ belong to V . Since |x∗∗(x∗)|=|x∗∗(x∗) − ξ(x∗∗) + ξ(x∗∗)|≥ |ξ(x∗∗)|−|x∗∗(x∗) − ξ(x∗∗)| >b+ − = b, either x∗ or −x∗ belongs to V ∩ S(x∗∗,b).

~~Now we prove the following result [92].~~

~~Theorem 13.3 (Cascales–Marciszewski–Raja) If E is a Banach space with the property (C), then for every bounded set H ⊂ E we have ck(H ) = k(H ).~~

Proof By Theorem 4.12, we know that ck(H ) ≤ k(H ). Then, if k(H ) = 0, the required equality holds. Therefore, it is enough to show that for each 0 b. Set ∗∗ ∗ ∗∗ ∗ S(x ,b):= {x ∈ BE : x (x )>b}. (13.1) 284 13 Corson’s Property (C) and Tightness

ω∗ By Lemma 13.1,wehave0∈ S(x∗∗,b) . By Theorem 13.2, there exists a count- ∗ able subset C ⊂ S(x∗∗,b) such that 0 ∈ convω C.AsS(x∗∗,b) is convex, there ω∗ ω∗ exists a countable set D ⊂ S(x∗∗,b) with 0 ∈ D . Since H is pseudometrizable ∗∗ in the pointwise topology on D, there exists a sequence (hn)n in H with hn → x ∗∗ ∗ on D.Ifh is any ω -cluster point of (hn)n,wehave h∗∗|D = x∗∗|D. ω∗ Consequently, h∗∗(x∗) = x∗∗(x∗)>bfor each x∗ ∈ D. Since 0 ∈ D , for ﬁxed arbitrary y ∈ E and >0 there exists x∗ ∈ D with |x∗(y)| <.This is applied to get h∗∗ − y ≥h∗∗(x∗) − x∗(y) > b − . ∗∗ ∗ ∗∗ Hence d(h ,E)≥ b for each ω -cluster point h of ϕ = (hn)n. Consequently,

~~ck(H ) ≥ d(clustE (ϕ), E) ≥ b.~~

~~Theorem 13.3 provides a large class of Banach spaces for which the numbers ck(H ) and k(H ) coincide; for example, every separable Banach space has this property.~~

~~13.2 The property (C) for Banach spaces C(X)~~

Let X be a topological space, and choose x ∈ X.Bythetightness of X at the point x we mean (and denote by t(x, X)) the smallest infinite cardinal m such that for each set A ⊂ X with x ∈ A there exists a subset B ⊂ A with |B|≤m such that x ∈ B. A topological space X has tightness m if m is the smallest infinite cardinal such that t(x, E) ≤ m for each x ∈ E. If K is a compact space, the dual C(K) is identified with M(K), the space of signed Radom measures on K of finite variation. Let M0(K) be the unit ball in M(K).ByP(K)we denote the set of probabilistic Radom measures on K endowed with the topology σ (P (K), C(K)). We need the following property; see [146, Problem 3.12.8 (a),(c),(f)].

~~Lemma 13.2 For a compact space K, the following hold: (i) t(P (K)) = t(M0(K)). (ii) If T ⊂ K is a closed Gδ-subset and x ∈ T , then t(x, T ) = t(x, K).~~

~~Proof (i) By the deﬁnition, t(P (K)) ≤ t(M0(K)).Nowset A := {(x, y) :|x|+|y|≤1}. Since the map~~

~~U : P(K)× P(K)× A → M0(K) 13.2 The property (C) for Banach spaces C(X) 285 deﬁned by U(μ,ν,x,y) := xμ − yν is a continuous surjection, we have~~

~~t(M0(K)) ≤ t(P (K)) by using [146, 3.12.8 (a),(f)]. For part (ii), we refer to [146, 3.12.8 (c)].~~

~~Motivated by the facts above, we show the following result (see [168, Theo- rem 2.3]); the proof uses some ideas from [331].~~

~~Theorem 13.4 (Frankiewicz–Plebanek–Ryll-Nardzewski) If K is a compact space with countable tightness, the space C(K) is realcompact in the weak topology.~~

Proof By Proposition 12.1, it is enough to show that if ξ ∈ (C(K))∗ is a functional that is σ(C(K), C(K))-continuous on each σ(C(K), C(K))-separable subspace, then ξ ∈ C(K). Deﬁne ϕ on K by ϕ(x) := ξ(δx), where δx is the Dirac measure at x. Since ξ is σ(C(K), C(K))-continuous on the subspace spanned on {δx : x ∈ S}, where S is an arbitrary separable subspace of K,themapϕ is continuous on S. Since K has countable tightness, the same argument used in the proof of Theorem 12.2 is applied to show that ϕ is continuous on K (i.e., ϕ ∈ C(K)). To show that ξ ∈ C(K), it is enough to prove that ξ(μ) = μ(ϕ) for every μ ∈ P(K), and hence we need to show that ϑ := ξ − ϕ = 0. Assume, by contradiction, that ϑ(μ) > 0forsome μ ∈ P(K). By the Radon–Nikodym theorem [288, Theorem 13.12], applied to the μ-continuous measure γ(B):= ϑ(μB ), where B ⊂ K is a Borel set and μB := μ|B, there exists a μ-measurable function g on K such that

ϑ(μB ) = gdμ B for each B.Fixa>0 and closed M ⊂{x ∈ K : g(x) ≥ a} with μ(M) > 0. If η := −1 (μ(M)) μM ,wehaveϑ(ηB ) ≥ aη(B) for all μ-measurable B contained in M. Observe that we may assume μ(M ∩ V)>0ifV ⊂ K is open with V ∩ M = ∅. Finally, there exist x ∈ M, a sequence (Un)n of open subsets of M such that for each neighborhood Ux of x there exists m ∈ N such that Um ⊂ Ux ; this can be checked in [374]. Set := −1 := { : ∈ N} ηn (η(Un)) ηUn , F ηn n . Then δx ∈ F,ϑ(ηn) ≥ a, and ϑ(δx) = 0. Since ϑ is σ(C(K), C(K))-continuous on F, we reach a contradiction.

In [337, Corollary 4.2], Pol showed that for compact K such that each μ ∈ P(K) is countably determined (i.e., there exists a countable family F of compact subsets of K such that μ(B) = sup{μ(F ) : F ⊂ B,F ∈ F} for every open B), the space C(K) has the property (C) if and only if P(K) has countable tightness. This result was generalized in [168, Theorem 3.2] to compact spaces K such that every μ ∈ P(K) is separable (i.e., the Banach space L1(μ) is separable). 286 13 Corson’s Property (C) and Tightness

~~Theorem 13.5 (Frankiewicz–Plebanek–Ryll-Nardzewski) Let K be a compact space such that every μ ∈ P(K) is separable. Then C(K) has the property (C) if and only if P(K)has countable tightness.~~

Proof Assume P(K) has countable tightness. If C(K) does not have the property (C), the space M0(K) does not have countable tightness because the set A considered in Theorem 13.2 (see that (ii) → (i)) is a subset of M0(K). Lemma 13.2 implies that P(K)does not have countable tightness, a contradiction. To prove the converse, assume that the space C(K) has the property (C). Assume also that there exists separable μ ∈ P(K) such that the tightness t(μ,P(K)) is uncountable. By the assumption on μ,let(fn)n be a sequence in C(K) that is dense in L1(μ).Fix U := {ν ∈ P(K): ν(fn) = μ(fn)}. n Applying Lemma 13.2 (ii) to the zero-set U, we note that t(μ, U) is uncountable. Hence there exists an uncountable family W ⊂ U such that μ ∈ W and μ/∈ N for each countable N ⊂ W.Fix>0. For each scalar t and each g ∈ C(K), the product tg ∈ C(K). Therefore, for each countable N ⊂ W, there are n ∈ N and g1,...,gn ∈ Cc(K) such that

N ∩ V(g1,...,gn, 3) =∅, (13.2) where V(g1,...,gn, 3) := {ν ∈ C(K) :|ν(gj ) − μ(gj )| < 3} 1≤j≤n is a σ(C(K), C(K))-neighborhood of μ.Forν ∈ P(K),theset T(ν):= {g ∈ : ≥ ≤ } =∅ C(K) ν(g) 2,μ(g) is convex, closed in C(K), and ν∈W T(ν) , since μ ∈ W. ⊂ We prove that for each countable N W the set ν∈N T(ν)is nonempty. This will yield a contradiction since we show that C(K) does not have the property (C). Fix a countable family N ⊂ W, and choose functions g1,g2,...,gn in C(K) satisfying (13.2). Next, choose functions fj1 ,...,fjn from the sequence (fn)n such that

~~| − | −1 fjl gl dμ < n K~~

= := | − | ≤ ∈ for each l 1, 2,...,n.Ifh 1≤l≤n fjl gl , we have μ(h) .Forν N, there exists 1 ≤ l ≤ n such that |ν(gl) − μ(gl)|≥3. Hence ≥| − |≥| − |−| − |≥ ν(h) ν(gl) ν(fjl ) ν(gl) μ(gl) ν(fjl ) μ(gl) −| − |≥ 3 μ(fjl ) μ(gl) 2. ∈ This shows that h ν∈N T(ν), proving that C(K) does not have the property (C). 13.2 The property (C) for Banach spaces C(X) 287

In the ﬁrst part of the proof of Theorem 13.5, we proved that the countable tightness of P(K)always yields the property (C) for C(K). To obtain the converse, one can ask about nonseparable measures in the space P(K). Fremlin [170] proved that under axioms MA and ¬ CH the following assertions are equivalent: (i) Every measure μ ∈ P(K)is separable. (ii) There does not exist a continuous surjection from K onto [0, 1]ω1 . This yields the following proposition.

~~Proposition 13.3 Under MA and ¬ CH, the following assertions are equivalent for a compact space K: (i) The space P(K)has countable tightness. (ii) The Banach space C(K) has the property (C).~~

Proof (i) ⇒ (ii): See the ﬁrst part of the proof of Theorem 13.5. ω (ii) → (i): By Proposition 13.2, we know t(K) ≤ℵ0. Since the space [0, 1] 1 does not have countable tightness, Fremlin’s result is applied to deduce that every measure in P(K)is separable and Theorem 13.5 applies.

From [168, Lemma 3.5], it follows that if K is a compact zero-dimensional space and C(K) is weakly Lindelöf, every measure μ ∈ P(K)is separable. This yields the following result of [168, Theorem 3.6].

~~Theorem 13.6 If K is a compact zero-dimensional space and C(K) is a weakly Lindelöf space, the space P(K)has countable tightness. Chapter 14 Fréchet–Urysohn Spaces and Groups~~

Abstract This chapter deals with topological (vector) spaces satisfying some sequential conditions. We study Fréchet–Urysohn space (i.e., spaces E such that for each A ⊂ E and each x ∈ A there exists a sequence in A converging to x). The main result states that every sequentially complete Fréchet–Urysohn lcs is a Baire space. Since every inﬁnite-dimensional Montel (DF )-space E is nonmetrizable and sequential, the following question arises: Is every Fréchet–Urysohn space in the class G metrizable?

~~14.1 Fréchet–Urysohn topological spaces~~

Recall that a topological space E is called Fréchet–Urysohn if for every subset A ⊂ E and every x ∈ A there exists a sequence from A converging to the point x. We say that a topological space E satisﬁes the Fréchet–Urysohn property if E is a Fréchet–Urysohn space. The Fréchet–Urysohn property is known to be highly non- multiplicative; the square of a compact Fréchet–Urysohn space need not be Fréchet– Urysohn; see [379]. Van Douwen [141] proved that the product of a metrizable space by a Fréchet–Urysohn space may not (even) be sequential. Recall that E is said to be sequential if every sequentially closed subset of E is closed. These concepts have been studied by topologists and analysts over the last half century (see, for example, [137], [144], [146], [169], [224], [229], [230], [316], [376], [141], [415], [435], [436]; see also the recent survey paper [365]). It is known that a Fréchet–Urysohn lcs may not be metrizable. Probably the ﬁrst example of such a space was presented in [39]. The following list of certain results provides many examples of a nonmetrizable Fréchet–Urysohn lcs. We refer to [179, Theorem 2], [27, Theorem II.3.7 and Theorem III.1.2], [90, Corollary 4.2], [373, Theorem 5.1], [204, Theorem 1, Theorem 2], [407] and [397].

Proposition 14.1 (i)(McCoy–Gerlits–Nagy) X is an ω-space (i.e., every open ω- cover of X has a countable ω-subcover) if and only if Cp(X) is Fréchet–Urysohn (for the deﬁnition of an ω-cover, see Proposition 9.19). (ii)(Pytkeev–Gerlits–Nagy) Cp(X) is Fréchet–Urysohn if and only if Cp(X) is sequential if and only if Cp(X) is a k-space (for the deﬁnition of a k-space, see the text before Lemma 6.8).

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 289 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_14, © Springer Science+Business Media, LLC 2011 290 14 Fréchet–Urysohn Spaces and Groups

(iii)(Arkhangel’skii–Pytkeev) If X is compact, Cp(X) is Fréchet–Urysohn if and only if X is scattered. (iv)(Pytkeev–Gerlits) Cc(X) is Fréchet–Urysohn if and only if Cc(X) is sequential if and only if Cc(X) is a k-space. (v)(Hernández–Mazón) If X is ﬁrst-countable, the space Cc(X) is Fréchet– Urysohn if and only if X is hemicompact. N (vi)(Arkhangel’skii–Tkachuk) Cp(X) is Fréchet–Urysohn for Fréchet–Urysohn Cp(X). There exist Fréchet–Urysohn spaces Cp(X) and Cp(Y ) such that Cp(X) × Cp(Y ) does not have countable tightness (Todorˇcevi´c).

Proposition 14.2 is an unpublished result of Morishita; see [226, Theorem 2] and also [24, Theorems 10.5, 10.7] and [23, Theorem II.7.16]. Proposition 14.2 was extended in [90, Corollary 4.2] to K-analytic spaces X. For the deﬁnition of a kR- space, see the text before Theorem 9.4.

~~Proposition 14.2 For a Cech-completeˇ and Lindelöf space X, the following assertions are equivalent: (i) X is scattered. (ii) Cp(X) is Fréchet–Urysohn. (iii) Cp(X) is a kR-space.~~

Proof (i)⇒(ii) holds for any Lindelöf space [27, Theorem II.7.16.]. (ii)⇒ (iii) is obvious. (iii) ⇒ (i): Assume X is not scattered. Then X contains a compact space Y that can be continuously mapped onto the interval [0, 1] (see [90, Theorem 4.1]) since every Cech-completeˇ and Lindelöf space is K-analytic. Let ϕY be the restriction map of Cp(X) into Cp(Y ) defined by ϕY (f ) = f |Y for any f ∈ Cp(X). Clearly, ϕY is Y continuous and open, and ϕY (Cp(X)) = Cp(Y ). Set W := Cp(Y )∩[0, 1] . Observe that W is not a kR-space. Let μ be a finite nonnegative nonatomic regular Borel : →[ ] = measure on Y . Define a map ψ W 0,μ(Y) by the formula ψ(f) Y fdμ for any f ∈ W . Note that ψ is not continuous at zero. Indeed, for every ε>0, f ∈ W , and ⊂ Y ,set

~~={g ∈ W :|f(x)− g(x)| <ε,x∈ }.~~

Assume that ψ is continuous at zero. Then there exists a ﬁnite subset ⊂ Y , ε>0, such that − ψ(<0,,ε>)⊂[0, 2 1μ(Y )). Since μ() = 0, by the regularity of μ, there exists a closed subset F in Y such that ∩ F =∅and μ(F ) ≥ 2−1μ(Y ). There exists a function f ∈ W such that f |F = 1 and f | = 0. Finally, ψ(f)≥ 2−1μ(Y ), a contradiction with f ∈< 0,,ε>. Now we show that ψ is kR-continuous (i.e., for every compact subset K of W , the map ψ|K is continuous). Y is compact, so the space Cp(Y ) is monolithic (i.e., the closure of any countable subset of Cp(Y ) is a space with a countable network [27, Theorem II.6.19], and has countable tightness by Proposition 9.9). If K is a 14.2 A few facts about Fréchet–Urysohn topological groups 291 compact subset of W ⊂ Cp(Y ), then it is also monolithic (see [27, Theorem II.6.5]) and clearly has countable tightness. We claim that K is a Fréchet–Urysohn space. Indeed, let A ⊂ K be a subset of K, and let x ∈ A (the closure in K). By the assumption, there exists a countable set B ⊂ A such that x ∈ B. Since K is monolithic, B has a countable network. As every compact space with a countable network is a space with a countable base, there exists in B a sequence that converges to x. Hence, to show that ψ|K is continuous, it is enough to prove that ψ|K is sequentially continuous. Let (xn)n be a sequence in

~~Y K ⊂[0, 1] ∩ Cp(Y ), and assume that xn → x in K. Therefore~~

~~0 xn(t) 1,xn(t) → x(t), t ∈ Y, n ∈ N.~~

By the Lebesque dominated convergence theorem, we have ψ(xn) → ψ(x). Finally, we show that Cp(Y ) is not a kR-space. Let r : R →[0, 1] beamap deﬁned by r(x) = 1forx 1, r(x) = x for 0 x 1 and r(x) = 0forx 0. Deﬁne a map : Cp(Y ) → W by (f) = rf for any f ∈ Cp(Y ). Then is the retraction. Since W is not a kR-space, the space Cp(Y ) is not a kR-space. Since ϕY is continuous and open, Cp(X) is not a kR-space.

~~14.2 A few facts about Fréchet–Urysohn topological groups~~

~~Recall the concept of the double sequence property;see[23], [22].~~

Deﬁnition 14.1 (α4) For any family {xm,n : (m, n) ∈ N × N}⊂X with limn xm,n = x ∈ X, m ∈ N, there exist a sequence (ik)k of distinct natural numbers and a se- = quence (jk)k of natural numbers such that limk xik,jk x.

Every Fréchet–Urysohn topological group satisﬁes the property (α4). This property fails for topological spaces in general [316, Theorem 4]. By [38, Lemma 3.3], a Fréchet–Urysohn tvs satisﬁes the following (stronger) property (see Lemma 14.1).

Deﬁnition 14.2 We shall say that X satisﬁes the property (as) if for any family {xm,n : (m, n) ∈ N × N}⊂X with limn xm,n = x ∈ X for m ∈ N there exist two strictly increasing sequences of natural numbers (ik)k and (jk)k such that = limk xik,jk x.

~~We need the following lemma from [94, Lemma 1.3].~~

~~Lemma 14.1 A Fréchet–Urysohn Hausdorff topological group X satisﬁes the property (as) and hence the property (α4) as well. 292 14 Fréchet–Urysohn Spaces and Groups~~

~~Proof By 0 we denote the neutral element of X. It is enough to show the property (as) for any family~~

{xm,n : (m, n) ∈ N × N}⊂X with limn xm,n = 0,m∈ N. Fix a sequence (am)m ⊂ X with limm am = 0 and am = 0form ∈ N (if such a sequence does not exist, then X is discrete and the conclusion is trivial). Set

~~ym,l := am + xm,l+m if am + xm,l+m = 0 and ym,l := am otherwise. Set~~

~~M := {ym,l : (m, l) ∈ N × N}.~~

Then 0 ∈ M. Note that 0 ∈ M.LetU and U0 be neighborhoods of zero with U + U ⊂ U0. Since limm am = 0, there is m ∈ N such that am ∈ U, and by limn xm,n = 0 there is l ∈ N such that xm,l+m ∈ U. Hence ym,l ∈ U + U ⊂ U0. Next, by the Fréchet–Urysohn property, and since 0 ∈ M, there exists a sequence (mk,lk)k = such that limk ymk,lk 0. Case 1. The sequence (lk)k is bounded. Taking a subsequence if necessary, we may assume that lk = r for some natural number r. Since

0 = lim ym ,l = lim ym ,r k k k k k = =∞ and ymk,r 0, we conclude that limk mk . Wemayalsoassumethatm1 < m2 <.... Assume ﬁrst that ={ ∈ N : = } N1 k ymk,r amk is inﬁnite. Set N ={p ,p ,...} with p

~~y = a + x + mqk ,lqk mqk mqk ,r mqk for k ∈ N. Since lim y = 0 and lim a = 0, we note that k mqk ,lqk k mqk~~

~~lim xm ,r+m = 0. k qk qk 14.2 A few facts about Fréchet–Urysohn topological groups 293~~

:= := + ∈ N Again set ik mqk and jk r mqk for k . Then there exist strictly increasing = sequences (ik)k and (jk)k such that limk xik,jk 0. Case 2. The sequence (lk)k is not bounded. We may assume that (lk)k is strictly increasing. Then limk mk =∞. Indeed, otherwise taking a subsequence if necessary, we may assume that mk = s for some s. The sequence (lk)k is strictly increas- = = = = ing, so we have limk xs,s+lk 0. From limk ys,lk 0, it follows that amk as 0, a contradiction with the choice of (am)m. Hence limk mk =∞. There exists a strictly increasing sequence (nk)k such that mn1

One of the interesting problems (due to Malyhin) concerning Fréchet–Urysohn groups asks whether it is consistent that every countable Fréchet–Urysohn topological group be metrizable [373];seealso[302] and [408] for some examples under various additional set-theoretic assumptions. Some possible approaches for attack- ing this problem have recently been provided in [67]. We show that under Martin’s axiom (MA) there exist nonmetrizable analytic (hence separable) Fréchet–Urysohn lcs’s. We observe that the Borel conjecture implies that separable Fréchet–Urysohn spaces Cp(X) are metrizable. On the other hand, Laver [265] proved that it is relatively consistent with the axioms of ZFC (i.e., Zermelo–Fraenkel axioms of set theory plus the axiom of choice) that the Borel conjecture is true. In fact, there exist many important classes of lcs for which the Fréchet–Urysohn property implies the metrizablity; we will see that every lcs E in the class G is metrizable if and only if E is Fréchet–Urysohn.

~~Example 14.1 There exists a σ -compact, and hence K-analytic, nonmetrizable Fréchet–Urysohn topological group that is not Baire.~~

Proof The direct sum X of the ℵ1 copies of the circle group (R/Z, +) is a σ - compact Fréchet–Urysohn group that is nonmetrizable; see [302, Example 1.2]. As- sume X is Baire. Then X is locally compact. As every locally compact Fréchet– Urysohn topological group is metrizable (Proposition 9.24), we reach a contradiction.

~~We also have the following example.~~

Example 14.2 The -product X of the compact group {0, 1}ω1 (i.e., the set of elements of the product {0, 1}ω1 with at most countably many nonzero coordinates) is Fréchet–Urysohn, locally compact and not K-analytic. Nevertheless, under CH it has a compact resolution. On the other hand, under MA +¬ CH the space X does not have a compact resolution. 294 14 Fréchet–Urysohn Spaces and Groups

Proof Note that X is Fréchet–Urysohn and locally compact (see [315, Theorem 2.1] and [42, Chapter IX, Exercise 17]). Note also that X is not Lindelöf. Indeed, the following open cover {Ut : t ∈[0,ω1)} of X does not admit a countable subcover, where Ut is the set of all points of X whose t-coordinate is 0. Since every K-analytic space is a Lindelöf space, X is not K-analytic. If CH is assumed, the space [0,ω1) N has a compact resolution {Aα : α ∈ N } swallowing compact sets (Proposition 3.9). N Deﬁne compact sets Kα for α ∈ N in X by the formula

Kα ={x = x(t) ∈ X : x(t) = 0,t/∈ Aα}. N Note that {Kα : α ∈ N } is a compact resolution swallowing compact sets of X. On the other hand, under MA +¬ CH, the space [0,ω1) does not have a compact resolution by Proposition 3.9, and the same property holds for X.

We provided examples of nonmetrizable Fréchet–Urysohn σ -compact topological groups. Example 14.3 extends Webb’s theorem [412, Theorem 5.7]. Webb proved that only ﬁnite-dimensional Montel (DF)-spaces are Fréchet–Urysohn. Re- call that every Fréchet–Urysohn Montel (DF)-space is a hemicompact tvs, and every locally compact tvs is ﬁnite-dimensional.

~~Example 14.3 Every Fréchet–Urysohn hemicompact topological group X is a locally compact Polish space.~~

Proof Let (Kn)n be an increasing sequence of compact sets covering X such that every compact set in X is contained in some Km. Note that X is locally compact. Indeed, it is enough to show that there exists n ∈ N such that Kn contains a neighborhood of the unit of X.LetF be a basis of neighborhoods of the unit of X.As- sume that no Kn contains an element of F. For every U ∈ F and n ∈ N, choose xU,n ∈ U \ Kn, and for each n ∈ N let

~~An ={xU,n : U ∈ F}.~~

Since 0 ∈ An for every n ∈ N, there exists a sequence (Um,n)m in F such that → →∞ := xm,n 0, m , where xm,n xUm,n,n. By Lemma 14.1, there exist a sequence N N → (nk)k of distinct numbers in and a sequence (mk)k in such that xmk,nk 0. { : ∈ N}∪{ } ∈ ∈ N As xmk,nk k 0 is contained in some Kp and xmk,nk / Knk for each k , we reach a contradiction. Hence X is a locally compact Fréchet–Urysohn group. By Proposition 9.24, we know that X is metrizable. Hence X is analytic. Since any analytic Baire topological group is a Polish space (Theorem 7.3), the proof is ﬁnished.

We complete this section with a simple characterization of Fréchet–Urysohn spaces Cc(X) over locally compact metric spaces X; this supplements Theo- rem 9.17. First we note the following fact; see also Theorem 14.3.

~~Lemma 14.2 If X = (X, d) is a metric space and Cc(X) is a Fréchet–Urysohn space, the space X is separable. 14.2 A few facts about Fréchet–Urysohn topological groups 295~~

Proof We may assume that X is noncompact, so there exists on X an equivalent unbounded metric. Let K (X) be the family of all compact sets in X.ForK ∈ K (X),letVK be an open neighborhood of K such that − d(x,K) < (max{1,δ(K)}) 1 for each x ∈ VK , where δ(A) means the diameter of A. Then −1 δ(VK ) ≤ δ(K) + 2(max{1,δ(K)}) . Let

~~fK : X →[0, max{1,δ(K)}] be a continuous function such that fK (x) = 0 for all x ∈ K and fK (y) = max{1,δ(K)} for each y ∈ X \ VK . Set~~

~~M := {fK : K ∈ K (X)}.~~

~~Then 0 ∈ M, where the closure is taken in Cc(X). By the assumption, there exists a bounded set B ⊆ M whose closure contains 0. Observe that~~

sup{δ(K) : fK ∈ B}=∞. Indeed, assume on the contrary that for some n ∈ N one has δ(K) ≤ n for all K ∈ K (X) for which fK ∈ B. Choose Q ∈ K (X) with δ(Q) > n + 2, and set − U[Q]:= f ∈ C(X) : sup |f(x)| < 2 1 . x∈Q

~~Since 0 ∈ B, there exists fK ∈ B ∩ U[Q],soQ ⊆ VK , and hence~~

−1 δ(Q) ≤ δ(VK ) ≤ δ(K) + 2(max{1,δ(K)}) ≤ n + 2, a contradiction. ∈ N ∈ K ∈ ≥ For every n , choose Kn (X) such that fKn B and δ(Kn) n.If ∈ ∈ N ∈ ≥ = there are y X and k such that y/VKn for each n k,wehavefKn (y) max{1,δ(Kn)}≥n for each n ≥ k, which leads to sup |f (y)| =+∞, f ∈B a contradiction with the boundedness of B. Therefore, for each x ∈ X, k ∈ N, there exists n ≥ k (where n depends on x) such that x ∈ V . Hence k k Knk ≤ { } −1 ≤ −1 d(x,Knk ) (max 1,δ(Knk ) ) nk . Since each compact metric space is separable, for each k ∈ N there is a countable set Ak in Knk dense in Knk . Choosing

xk ∈ Ak ∩ B(x,2/nk), 296 14 Fréchet–Urysohn Spaces and Groups where B(x,2/nk) stands for the open ball with center at x and radius 2/nk,wesee → ∞ that xk x in X. Hence the countable set k=1 Ak is dense in X, and then X is separable.

~~Now we prove the following proposition.~~

Proposition 14.3 If X is a locally compact metric space, the following assertions are equivalent: (i) Cc(X) is a Fréchet space. (ii) Cc(X) is Fréchet–Urysohn. (iii) Cc(X) has countable tightness. (iv) X is σ -compact.

Proof Since every locally compact separable metric space is hemicompact, Lemma 14.2 proves (i) ⇔ (ii). The implication (iv) ⇒ (i) follows from the fact that any metric σ -compact space is separable and hence hemicompact. To complete the proof, it sufﬁces to show the implication (iii) ⇒ (iv): For every K ∈ K (X), there exists an open set UK with the compact closure such that

~~K ⊆ UK ⊆ UK .~~

Choose fK ∈ C(X) such that fK (x) = 0 for each x ∈ K and fK (x) = 1 for all x ∈ X \ UK . Then 0 belongs to the closure (in Cc(X))of{fK : K ∈ K (X)}.By the assumption, there exists a sequence Kn ∈ K (X), n ∈ N, such that the closure { : ∈ N} ∈ K ∈ N of fKn n contains 0. Hence, for arbitrary Q (X), there is j with | | −1 ⊆ supx∈Q fKj (x) < 2 , so that Q UKj . This shows that X is σ -compact.

~~14.3 Sequentially complete Fréchet–Urysohn spaces are Baire~~

~~In this section, we prove the following main result from [227].~~

~~Theorem 14.1 Every sequentially complete Fréchet–Urysohn lcs is a Baire space.~~

To prove Theorem 14.1, we ﬁrst show Lemma 14.3 due to Burzyk [74]. Recall that a sequence (xn)n in a topological additive group X is called a K-sequence if every subsequence of (xn)n contains a subsequence (yn)n such that the series n yn converges in X;see[11] and also [75] and [328]. Clearly, if X is a metric and complete additive topological group, every sequence in X converging to zero is a K-sequence.

Lemma 14.3 Let (Xk)k be an increasing sequence of closed subsets of a topological additive group X covering X. Let (xn)n be a K-sequence in X. Then there exists ∈ N ∈ +{− : ⊂{ }} ∈ N m such that xn Xm k∈A xk A 1, 2,...m for every n . 14.3 Sequentially complete Fréchet–Urysohn spaces are Baire 297

Proof Suppose the conclusion fails. Since (Xn)n is increasing, and subsequences ∈ N ∈ + of K-sequences are K-sequences, for each n there exists xn+1 / Xn {− : ⊂{ }} = k∈A xk A 1, 2,...n , where x1 0. Set Gn+1 := Xn + − xk : A ⊂{1, 2,...n} k∈A and G1 := {0}.ThesetsGn areclosedinX, so for each n ∈ N there exist a continuous pseudonorm qn on X and εn > 0 such that

~~inf qn(xn − z) > εn. (14.1) z∈Gn~~

Since xn → 0, there exists a sequence (kn)n in N such that − −n−m qn( xkm )<2 εn for allm ≥ n and n, m ∈ N. Then there exists a subsequence (sn)n of (kn)n such = ∈ ∈ N that n xsn x X. The sequence (Xn)n covers X. There exists m such that ∈ = − m−1 ∈ x Xsm−1 .Ifu x n=1 xsn ,wehaveu Gsm and ∞ − =− xsm u xsm . n=m+1 − ≤ We conclude that qsm (xsm u) εsm , providing a contradiction with (14.1).

Corollary 14.1 Let (En)n be an increasing sequence of closed subsets of a Fréchet–Urysohn sequentially complete lcs E covering E. If (xn)n is a sequence in E such that xn → 0, there is a strictly increasing sequence (nk)k in N and m ∈ N ∈ +{− : ⊂{ }} ∈ N such that xnk Enm j∈A xnj A 1, 2,...m for each k .

Proof By τ we denote the original topology of E. For each k,n ∈ N,setxn,k := kxn. Since for any ﬁxed k ∈ N one has xn,k → 0forn →∞, by Lemma 14.1 there exists → →∞ a strictly increasing sequence (nk)k such that kxnk 0ifk .LetB be the { : ∈ N} closed, absolutely convex hull of the bounded set kxnk k . Then the linear span EB of B endowed with the Minkowski functional norm

−1 xB := inf{ε>0 : ε x ∈ B} is a Banach space (see [246, 20.11(2)]), and τ|EB ≤ τB , where τB is the topology → generated by the norm x B . Clearly, xnk 0inτB . Then each subsequence of (xnk )k contains a subsequence (zk)k whose series k zk converges in τB and hence = ∩ ∩ in τ . Since EB k Enk EB , and the sets Enk EB are closed in τB , we apply Lemma 14.3 to get m ∈ N such that for each k ∈ N we have ∈ ∩ + − : ⊂{ } xnk Enm EB xnj A 1, 2,...m . j∈A 298 14 Fréchet–Urysohn Spaces and Groups

~~We also need the following two additional lemmas found in [74].~~

Lemma 14.4 Let X be a Fréchet–Urysohn additive topological group. Let (Xn)n be a decreasing sequence of dense subsets of X. Then there exists a sequence (xk)k in X such that xk → 0 and xk ∈ Xk for each k ∈ N.

Proof For each n ∈ N, there exists a sequence (xn,m)m in Xn such that xn,m → 0 if m →∞. By Lemma 14.1, there exist two strictly increasing sequences (nk)k, N → →∞ ∈ ⊂ (mk)k,in such that xnk,mk 0ifk . Clearly, xnk,mk Xnk Xk for each ∈ N := ∈ N k . To complete the proof, it is enough to set xk xnk,mk for all k .

Lemma 14.5 Let X be a Fréchet–Urysohn additive topological group. Let (Xn)n be an increasing sequence of closed subsets of X such that int Xn =∅for each ∈ N N n . Then there exists a strictly increasing sequence (tn)n in and a sequence → ∈ +{− : ⊂{ }} ∈ N xn 0 in X such that xn / Xtn k∈A xk A 1, 2,...n for all n .

Proof Take a sequence (x1,j )j in X such that x1,j → 0ifj →∞. Since the sets W1,j := Xj + − x1,n : A ⊂{(1,p), 1 ≤ p ≤ n} (1,n)∈A are closed with the empty interior, the complements X \ W1,j compose a decreasing sequence of open, dense subsets in X. By a simple induction, applying Lemma 14.4, we construct a matrix (xi,j )i,j in X such that

~~xi,j → 0,j→∞,i∈ N, (14.2)~~

xi,j ∈/ Xj + − xn,m : A ⊂{(k, l) : 1 ≤ k ≤ i, 1 ≤ l ≤ j} , (14.3) (n,m)∈A for i ≥ 2,j∈ N. By Lemma 14.1, there exist two strictly increasing sequences N → →∞ (tn)n, (ln)n in such that xln,tn 0ifn .By(14.3), we conclude that ∈ + − : ⊂{ } xln,tn / Xtn xlk,tk A 1, 2,...n k∈A ∈ N := ∈ N for each n . Finally, it is enough to set xn xln,tn for each n .

~~Now we are ready to prove Theorem 14.1.~~

Proof Suppose E is not a Baire space. By Theorem 2.3, there exists in E an absorbing closed and balanced set B with the empty interior. For each n ∈ N,set En := nB. Then the sequence (En)n of closed sets with the empty interior is increasing and covers the whole space E.Let(xn)n and (tn)n be sequences in E and 14.4 Three-space property for Fréchet–Urysohn spaces 299

N, respectively, constructed in Lemma 14.5 (i.e., ∈ + − : ⊂{ } xn / Etn xk A 1, 2,...n (14.4) k∈A ∈ N → := ∈ N for each n and xn 0). Set Wn Etn for each n . Clearly, the sequence (Wn)n covers E. From Lemma 14.1, it follows that there exists a strictly increasing sequence (nk)k in N, m ∈ N, such that ∈ + − : ⊂{ } xnk Wnm xnj A 1, 2,...m j∈A ∈ N ∈ +{− : ⊂{ }} for each k . This yields xnm Wnm j∈A xnj A 1, 2,...m , which contradicts (14.4). Hence E is a Baire space.

~~Theorem 14.1 is applied to get the following classical theorem.~~

~~Theorem 14.2Let {Et : t ∈ T } be a family of Fréchet spaces. Then the topological := product E t∈T Et is a Baire space.~~

Proof Since the 0-product E0 of E is a Fréchet–Urysohn, sequentially complete subspace of E, by Theorem 14.1 the space E0, endowed with the topology induced from E, is a Baire space. E0 is dense in E. Therefore E is a Baire space.

~~We complete this section with the following application of Theorem 14.1 above.~~

~~Corollary 14.2 Let X be a Lindelöf P-space. Then Cp(X) is a bornological Baire space.~~

Proof Since X is a Lindelöf P-space, each ﬁnite product Xn is a Lindelöf space. This implies that X is an ω-space; see [179, Proposition, pp. 156–157]. Indeed, let U be an open ω-cover of X. Then U n := {An : A ∈ U } is an ω-cover of Xn for ∈ N U ⊂ U U n n ∈ N eachn . Note that if n is countable and n covers X for each n ,the U U set n n is a countable ω-subcover of . Now, by Proposition 14.1(i), the space Cp(X) is Fréchet–Urysohn. Since X is a P-space, Cp(X) is sequentially complete; see [213, Theorem 3.6.7]. We apply Theorem 14.1 to deduce that Cp(X) is a Baire space. The proof is completed since every Fréchet–Urysohn lcs is bornological by Lemma 14.6 below.

~~14.4 Three-space property for Fréchet–Urysohn spaces~~

Todorceviˇ c[´ 405] constructed two Fréchet–Urysohn spaces Cp(X) and Cp(Y ) whose product is not Fréchet–Urysohn as having uncountable tightness. This implies that the Fréchet–Urysohn property is not a three-space property (see also Ex- ample 14.4 below). 300 14 Fréchet–Urysohn Spaces and Groups

In [94], it was proved that the product of a first-countable space X by a Fréchet– Urysohn topological group Y is a Fréchet–Urysohn space. Michael [291] proved that, if E is a first-countable topological space and F is a Fréchet–Urysohn topological space with the property (as) in Definition 14.2, the product space E × F is Fréchet–Urysohn. For a tvs, we have the following proposition.

~~Proposition 14.4 If F is a closed metrizable subspace of a tvs E such that the quotient E/F is a Fréchet–Urysohn space, the space E is a Fréchet–Urysohn space.~~

Proof Since F is a metrizable subspace of E, there exists a decreasing sequence (Vn)n of neighborhoods of zero in E such that Vn+1 + Vn+1 ⊂ Vn for all n ∈ N and the sequence (Vn ∩ F)n is a basis of neighborhoods of zero in F . To prove that E is Fréchet–Urysohn, it is enough to show that if X ⊂ E is an arbitrary set and 0 ∈ X, there exists a sequence in X converging to 0. Let Q : E → E/F be the quotient map. Let U (E) be the set of all neighborhoods of zero in E. Note that for each neighborhood of zero U ∈ U (E) and each n ∈ N there exists xU,n ∈ X ∩ U ∩ Vn. Hence Q(0) ∈ An, where

~~An := {Q(xU,n) : U ∈ U (E)}.~~

~~Since E/F is a Fréchet–Urysohn tvs, for each n ∈ N there exists a sequence (Uk(n))k in U (E) such that~~

~~lim Q(xU ,n) = Q(0). k k(n)~~

By the property (as) (Lemma 14.1), there exist two increasing sequences (kp)p and (np)p in N such that → →∞ Q(xUkp(np),np ) Q(0) if np . := ∈ N ∈ Set up xUkp(np),np for each p . Fix a balanced neighborhood of zero W U (E). Then there exist n ∈ N such that (Vn + Vn) ∩ F ⊂ W and m>nsuch that

[up + (W ∩ Vn)]∩F = ∅ for p>m. Therefore, there exist elements y ∈ W ∩Vn, u ∈ F , such that up +y = u. ∈ ⊂ ⊂ ∈ = + ∈ + ∩ Since up Vnp Vp Vn and y Vn, we deduce that u up y (Vn Vn) F ⊂ W. Therefore up = u − y ∈ W + W . This implies that (up)p is a sequence in X that converges to 0 in E. Hence E is Fréchet–Urysohn.

~~This yields the following result of Michael [291].~~

~~Corollary 14.3 (Michael) Let E be a Fréchet–Urysohn tvs and F a metrizable tvs. Then the product E × F is a Fréchet–Urysohn tvs.~~

We present another example showing that the Fréchet–Urysohn property is not a three-space property. We need the following two additional lemmas; see [230, 14.4 Three-space property for Fréchet–Urysohn spaces 301

~~Theorem 3.1] and [343, Lemma 1.4]. For Lemma 14.7, we omit the proof and refer to [343, Lemma 1.4].~~

Lemma 14.6 Let E be a Fréchet–Urysohn lcs. Let (An)n be an increasing bornivorous sequence in E. Then there exists m ∈ N such that Am is a neighborhood of zero in E. Hence E is b-Baire-like and bornological.

Proof Assume that none of the sets An is a neighborhood of zero in E.ByU (E) we denote the set of all absolutely convex neighborhoods of zero in E. Then, for each U ∈ U (E), n ∈ N, there exists xU,n ∈ U \ nAn. For each n ∈ N,set

~~Bn := {xU,n : U ∈ U (E)}.~~

Then 0 ∈ Bn for each n ∈ N. Since E is Fréchet–Urysohn, for each n ∈ N there U → ∈ N exists a sequence (Un(k))k in (E) such that xUn(k),n 0 for each n .By Lemma 14.1, every Fréchet–Urysohn lcs satisﬁes the condition (as) from Deﬁni- tion 14.2. Hence there exist strictly increasing sequences (kp)p and (np)p such that → →∞ xUnp(kp),np 0ifnp . On the other hand, the bounded set { : ∈ N} xUnp(kp),np p ∈ N is not included in any set npAnp for p , a contradiction, since (An)n is an increasing bornivorous sequence.

Lemma 14.7 Let H be a dense hyperplane in the Banach space 2. Fix y ∈ 2 \ H . Let G be a tvs. Let Z ⊂ G be a dense hyperplane. Fix x ∈ G \ Z. Then there exist an lcs E and a closed subspace F ⊂ E such that F is isomorphic to the space Z, the quotient space E/F is isomorphic to the quotient space 2/[y], and E has a quotient isomorphic to the space G/[x], where [x] means span{x}.

~~Now we are ready to show the following example.~~

Example 14.4 The Fréchet–Urysohn property is not a three-space property. RR Proof Let Z be the 0-product of the space at the point 0. The space Z is a Fréchet–Urysohn lcs. Fix x ∈ RR \ Z. The space G := Z +[x] is not a Fréchet– Urysohn space. Indeed, by Proposition 14.6, it is enough to show that G is not bornological in the induced topology: Let f : G → R be a linear functional on G deﬁned by f(y + λx) := λ, y ∈ Z, and λ ∈ R. Observe that Z = ker(f ) := {y ∈ G : f(y)= 0}. Hence, as Z is dense and sequentially closed in RR,ker(f ) is dense and sequentially closed in G. Consequently, f is discontinuous and sequentially continuous (hence bounded). Indeed, let A be a countable subset of R. Set ZA := {(xi) ∈ Z : xi = 0,i ∈ R \ A}. Then the restriction of f to ZA +[x] is continuous (since its kernel is the closed set ZA). This shows that G is not bornological. Next, we apply Proposition 14.4 to deduce that the quotient space G/[x] is not Fréchet– Urysohn. Let H be a dense hyperplane in the Banach space 2. Choose y ∈ 2 \ H. 302 14 Fréchet–Urysohn Spaces and Groups

By Lemma 14.7, there exists an lcs E containing a closed subspace F such that F is isomorphic to the Fréchet–Urysohn space Z, the quotient E/F is isomorphic to the Banach space 2/[y], and E has a quotient isomorphic to the quotient G/[x] (which is not Fréchet–Urysohn). This implies that E is not Fréchet–Urysohn since any Hausdorff quotient of a Fréchet–Urysohn tvs is Fréchet–Urysohn.

~~14.5 Topological vector spaces with bounded tightness~~

There is another interesting tightness-type condition, formally weaker than the Fréchet–Urysohn property. Let E be a tvs. If for every subset A of E, and x ∈ A ⊆ E, there is a bounded set B ⊆ A such that x ∈ B, the space E is said to have bounded tightness [87], [153]. The concept of the bounded tightness was formally deﬁned in [153] and then used in [87] for studying the weak topology of normed spaces. Since every Fréchet– Urysohn tvs has bounded tightness, it is natural to ask about the converse implication. It turns out that the following general result [235] holds.

Theorem 14.3 (Kakol–López-Pellicer–Todd) ˛ For a tvs E (not necessarily Haus- dorff), the following assertions are equivalent: (1) E is Fréchet–Urysohn. (2) For every subset A of E such that 0 ∈ A, there exists a bounded subset B of A such that 0 ∈ B. ∈ (3) For any sequence (An)n of subsets of E, each with 0 An, there exists a ⊂ ∈ N ∈ sequence Bn An, n , such that n Bn is bounded and 0 n≤k Bk for each n ∈ N.

Proof (1) ⇒ (2) is clear. (2) ⇒ (3): It is obvious that (3) holds if 0 ∈ An for inﬁnitely many n or if {0}=E. Therefore, we assume that 0 ∈ An\An, for each n ∈ N, and that there exists a null sequence (xn)n in E\{0}. For each n ∈ N, there exists a closed neighborhood Un of zero such that 0 ∈/ Un + xn. Let each Cn = Un ∩ An. Clearly, zero is in each Cn\Cn but not in the set

A := (Cn + xn) . n We claim that 0 ∈ A. Indeed, for U an open neighborhood of zero, there exist k ∈ N with xk ∈ U and V a neighborhood of zero with V + xk ⊂ U. Since there is y ∈ V ∩ Ck, we have y + xk ∈ U ∩ A. Thus 0 ∈ A\A. By the assumption, there is B ⊂ A with B bounded and 0 ∈ B. There exist subsets

~~Bn ⊂ Cn = Un ∩ An = + such that B n(Bn xn). By the construction, zero does not belong to the closed sets~~

~~(Uk + xk) . k~~

~~0 ∈ (Bk + xk), m≤k we see that there exist k ≥ m and y ∈ Bk with y + xk ∈ V .From~~

y ∈ V − xk ⊂ V − V ⊂ W ∈ N it follows that, for each n ,thesetW meets n≤k Bk. As any neighborhood of zero contains W and V as above, zero is in the closure of each n≤k Bk. We claim that n Bn is bounded. Indeed, as

~~B = (Bn + xn) n and D := {xn : n ∈ N} are bounded, and since~~

Bn ⊂ (Bn + xn) −{xm : m ∈ N}=B − D, n n the set n Bn is bounded, too. (3) ⇒ (1): Assume that 0 ∈ A, and set An := nA, for eachn ∈ N. Since 0 ∈ An ∈ N ⊂ for each n , there exist Bn An as in (3), so each n≤k Bk is nonempty N and consequently there exists a strictly increasing sequence (nk)k in with Bnk ∈ N ∈ nonempty. For each k , select yk Bnk . There exists a sequence (ak)k in A such that yk = nkak for each k ∈ N. As the sequence (nk)k is strictly increasing and (yk)k = (nkak)k is bounded, the sequence (ak)k in A converges to zero in E. Chapter 15 Sequential Properties in the Class G

Abstract In this chapter, we prove that an lcs in the class G is metrizable if and only if E is b-Baire-like if and only if E is Fréchet–Urysohn. Consequently, no proper (LB)-space is Fréchet–Urysohn. We prove that if a (DF )-or(LM)-space E is sequential, then E is either metrizable or Montel (DF ). We distinguish a variant − of the property C3 (due to Webb), called property C3 (i.e., sequential closure of any vector subspace is sequentially closed), and characterize both (DF )-spaces and − × (LF )-spaces with the property C3 as being of the form M, ϕ,orM ϕ, where M is metrizable.

~~15.1 Fréchet–Urysohn spaces are metrizable in the class G~~

~~In this section, we prove that every Fréchet–Urysohn lcs in the class G is metrizable; see [89]. Several applications will be provided. Let us start with the following lemma.~~

N Lemma 15.1 Let E be an lcs in the class G. Let {Aα: α ∈ N } be a G- = ∈ NN = { : = ∈ representation of E. For α (nk) , set Cn1...nk Aβ β (mk) NN,n = m ,j= 1, 2,...k} for k ∈ N. Then the sequence of polars C◦ ⊂ j j n1 C◦ ⊂···⊂C◦ ⊂··· is bornivorous in E. n1,n2 n1,n2,...,nk Proof Assume, by the contradiction, that there exists a bounded set B in E such that B ⊂ kC◦ for every k ∈ N. Then, for every k ∈ N there exists x ∈ B such n1...nk k that k−1x ∈ C◦ . Therefore, for every k ∈ N there exists f ∈ C such that k n1...nk k n1...nk −1 |fk(k xk)| > 1. ∈ N = k ∈ NN ∈ = k For every k , there exists βk (mn)n such that fk Aβk , where nj mj = = { k : ∈ N} ∈ N = ∈ NN for j 1, 2,...k. Next, deﬁne an max mn k ,n and γ (an) . ≥ ∈ N ⊂ ∈ ∈ N Since γ βk for k ,wehaveAβk Aγ ,sofk Aγ for all k . Finally, since by the deﬁnition any sequence in Aγ is equicontinuous, we note that the sequence (fk)k is equicontinuous. Hence the sequence (fk)k is uniformly bounded on bounded sets in E, including B, a contradiction.

~~We also need the following lemma.~~

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 305 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_15, © Springer Science+Business Media, LLC 2011 306 15 Sequential Properties in the Class G

Lemma 15.2 For a quasibarrelled space E, the following statements are equivalent: (i) E is in G. F := { : (ii) There is a family of absolutely convex closed subsets Dn1,n2,...,nk k,n1,n2,...,nk ∈ N} satisfying: ⊂ ≤ = (a) Dm1,m2,...,mk Dn1,n2,...,nk , whenever ni mi , i 1, 2,...,k. = ∈ NN ⊂ ⊂···⊂ ⊂··· (b) For every α (nk) , we have Dn1 Dn1,n2 Dn1,n2,...,nk and the sequence is bornivorous. := = ∈ NN { : ∈ NN} (c) If Uα k Dn1,n2,...,nk , α (nk)k , then Uα α is a basis of neighborhoods of the origin in E. N (iii) E has a G-basis (i.e., a basis of neighborhoods of zero {Uα : α ∈ N } satis- N fying the decreasing condition Uβ ⊂ Uα whenever α ≤ β in N ). (iv) The strong dual (E,β(E,E))is a quasi-(LB)-space.

N Proof (i) ⇒ (ii): Fix a G-representation {Lα : α ∈ N } of E. Since E is quasi- := ◦◦ barrelled, each set Lα is equicontinuous. Hence each set Mα Lα is weakly compact and β(E ,E)-bounded. This shows that each Mα is a β(E ,E)-Banach N disc. Therefore, {Mα : α ∈ N } is a resolution on E consisting of Banach discs in (E,β(E,E)). Hence the strong dual (E,β(E,E))is a quasi-(LB)-space. N By Theorem 3.5, there is a resolution {Aα : α ∈ N } in (E ,β(E ,E))consisting of Banach disc such that for every β(E,E)-Banach disc B ⊂ E there is α ∈ NN with B ⊂ Aα. The quasibarrelledness of E ensures that each set Aα is equicon- N tinuous, and then the G-representation {Aα : α ∈ N } is a fundamental family of { ◦ : ∈ NN} equicontinuous subsets of E . Hence the family of polars Aα α is a basis of neighborhoods of zero in E. ∈ N For k,n1,n2,...,nk , as usual, deﬁne Cn1,n2,...,nk and set

~~D := C◦ . n1,n2,...,nk n1,n2,...,nk~~

~~N For each α = (nk) ∈ N ,wehave ∞ σ(E,E ) ∞ ◦ := ⊂ ⊂ ◦ Vα Dn1,n2,...,nk Cn1,n2,...,nk Aα. k=1 k=1~~

~~Each set Vα is a closed, absolutely convex and bornivorous set by Lemma 15.1. Then Vα is a neighborhood of zero. By Proposition 2.13, we deduce that, for every ε>0, we have~~

~~ ∞ σ(E,E ) ∞ = ⊂ + = + Vα Dn1,n2,...,nk (1 ε) Dn1,n2,...,nk (1 ε)Uα. k=1 k=1~~

~~N Thus {Uα : α ∈ N } is a G-basis of neighborhoods of zero in E. (ii) ⇒ (iii) is trivial. 15.1 Fréchet–Urysohn spaces are metrizable in the class G 307~~

⇒ { ◦ : ∈ NN} (iii) (i): The family of polars Uα α is a G-representation of the space E. ⇒ { : ∈ NN} ◦ (ii) (iv): If Uα α is a G-basis in E,thesetsUα provide a quasi-(LB)- space representation for (E,β(E,E)). N (iv) ⇒ (ii): Let {Aα : α ∈ N } be a quasi-(LB) representation for (E ,β(E ,E)). Since E is quasibarrelled, each set Aα is equicontinuous. Hence E is in the class G and (i) ⇒ (ii) applies.

~~Now we are ready to prove Theorem 15.1. This result, due to Cascales, Kakol ˛ and Saxon [89], generalizes parts of [224, Theorem 5.1], [229, Theorem 2.1] and [310, Theorem 3].~~

~~Theorem 15.1 (Cascales–Kakol–Saxon) ˛ For an lcs E in G, the following statements are equivalent: (i) E is metrizable. (ii) E is Fréchet–Urysohn. (iii) E is b-Baire-like.~~

~~Proof (i) ⇒ (ii) is obvious. (ii) ⇒ (iii): See Lemma 14.6. (iii) ⇒ (i): If E is b-Baire-like, E is a quasibarrelled space, and therefore we can obtain a countable family~~

F := { : ∈ N} Dn1,n2,...,nk k,n1,n2,...,nk as in the proof of Lemma 15.2. ⊂ ⊂ ··· ⊂ ⊂ ··· Since the sequence Dn1 Dn1,n2 Dn1,n2,...,nk is bornivorous, = ∈ NN = ∞ for every α (nk) we have E k=1 kDn1,n2,...,nk and, again, since E is ∈ N b-Baire-like, some Dn1,n2,...,nm is a neighborhood of zero for certain m . Thus, by Lemma 15.2, the family

~~U := { ∈ F : } Dn1,n2,...,nk Dn1,n2,...,nk is neighborhood of 0 is a countable basis of neighborhoods of zero for E.~~

{ : ∈ } Corollary 15.1 Let Et t T be a family of nonzero lcs. If T is uncountable, the product t Et does not belong to the class G. Proof Assume that t∈T Et is in the class G, and T is an uncountable set. Then RA t∈T Et contains a subspace of the form for some uncountable set A. Clearly, RA is a Baire space in the class G. By Theorem 15.1, we deduce that RA is metrizable, a contradiction.

~~By Proposition 12.2, we know that Cp(X) belongs to the class G if and only if X is countable. 308 15 Sequential Properties in the Class G~~

~~Corollary 15.2 The following assertions are equivalent for a completely regular Hausdorff space X: (i) X is countable. (ii) Cp(X) is an (LM)-space. (iii) Cp(X) admits a G-basis.~~

Proof (i) ⇒ (ii) is clear. (ii) ⇒ (i): Since an (LM)-space belongs to the class G, we apply Proposition 12.2. (i) ⇒ (iii) is clear. (iii) ⇒ (i): The space Cp(X) is always quasibarrelled by [213, Corollary 11.7.3, Theorem 2]. We apply Lemma 15.2, and then again Proposition 12.2.

N Recall again that a bounded resolution {Kα : α ∈ N } in a tvs E is bornivorous if every bounded set in E is contained in some Kα.Clearly,every(DF )-space admits a bornivorous bounded resolution. Also, every regular (LM)-space admits a bornivorous bounded resolution. Indeed, let E be an (LM)-space. Let (En)n be a ∈ N n deﬁning sequence of E of metrizable lcs. For each n ,let(Uk )k be a countable basis of absolutely convex neighborhoods of zero in En such that

n ⊂ n+1 n ⊂ n Uk Uk , 2Uk+1 Uk , ∈ N = ∈ NN := n1 { : ∈ NN} for each k . For each α (nk) ,setKα k nkUk . Then Kα α ∈ ∈ N ∈ is a bounded resolution. Indeed, if x E, there exists r1 such that x Er1 . ∈ N ∈ N ∈ r1 = ∈ NN Hence for each k there exists mk such that x mkUk . Set α (nk) = { } = ≥ { : ∈ NN} with n1 max r1,m1 and nk mk for all k 2. Clearly, Kα α is a reso- n1 lution on E, and k nkUk is bounded in En1 and hence also in E. If additionally E is regular (i.e., for every bounded set B in E there exists m1 ∈ N such that B ∈ N ∈ N is contained and bounded in Em1 ), then for each k there exists nk such ⊂ m1 = ∈ NN ⊂ that B k nkUk . This yields a sequence α (nk) such that B Kα.The following corollary follows from Theorem 15.1.

Corollary 15.3 If an lcs E has a bornivorous bounded resolution, the strong dual (E,β(E,E))has a G-basis. For an lcs E, the strong dual (E,β(E,E))is metrizable if and only if (E,β(E,E)) is Fréchet–Urysohn and E admits a bornivorous bounded resolution.

Proof If (E,β(E,E)) is metrizable, it is Fréchet–Urysohn and E admits a bornivorous bounded resolution. Now assume E admits a bornivorous bounded reso- { : ∈ NN} ◦ lution Kα α . The polars Kα of the sets Kα in E form a basis of neigh- ◦◦ borhoods of zero for β(E ,E). Since the sets Kα in E compose a resolution consisting of equicontinuous sets covering E, the space (E,β(E,E))belongs to the class G.If(E,β(E,E))is Fréchet–Urysohn, Theorem 15.1 yields the metrizability of (E,β(E,E)). 15.1 Fréchet–Urysohn spaces are metrizable in the class G 309

Corollary 15.4 The spaces D(Ω) of the distributions and A(Ω), of the real analytic functions for an open set Ω ⊂ RN are not Fréchet–Urysohn; they have countable tightness both for the original and the weak topologies.

Proof Since D(Ω) is nonmetrizable and quasibarrelled, and is the strong dual of a complete (hence regular) (LF )-space D(Ω) of the test functions (see [288]), we apply Corollary 15.3 and Theorem 12.3. The same argument can be used for the space A(Ω) via [125, Theorem 1.7 and Proposition 1.7].

~~The following corollary, due to Cascales, Kakol ˛ and Saxon [89], extends [310, Theorem 3] since every (LF )-space is barrelled and belongs to the class G.~~

Corollary 15.5 For a barrelled lcs in the class G, the following conditions are equivalent: (i) E is metrizable. (ii) E is Fréchet–Urysohn. (iii) E is Baire-like. (iv) E does not contain ϕ (i.e., an ℵ0-dimensional vector space with the ﬁnest locally convex topology).

Proof Since every barrelled b-Baire-like space is Baire-like (see Proposition 2.12), the conditions (i), (ii) and (iii) are equivalent by Theorem 15.1. (iii) ⇒ (iv): By Theorem 15.1, the space E is metrizable. Since ϕ is a nonmetrizable (LF )-space, (iv) holds. (iv) ⇒ (i): Every barrelled space not containing ϕ is Baire-like by Corol- lary 2.4.

~~Since the completion of an lcs E ∈ G belongs to the class G and the completion of a quasibarrelled space is barrelled, Corollary 15.5 yields the following corollary.~~

~~Corollary 15.6 A quasibarrelled lcs in the class G is metrizable if and only if its completion does not contain ϕ.~~

We know already that every Fréchet–Urysohn lcs in the class G is metrizable and that only metrizable spaces Cp(X) belong to the class G. On the other hand, under (MA)+¬(CH) there exist nonmetrizable Fréchet–Urysohn spaces Cp(X). Indeed, by Proposition 14.1 and [284, Theorem 1], the space Cp(X) is Fréchet–Urysohn if and only if X has the γ -property (i.e., if for any open cover R of X such that any finite subset of X is contained in a member of R there exists a countable infinite subfamily R of R such that any element of X lies in all but finitely many members of R). Gerlits and Nagy [179] showed (under MA) that every subset of R of the car- ℵ dinality smaller than 2 0 has the γ -property. Hence, under MA+¬(CH), there are uncountable γ -subsets Y of R. Thus, for such Y the space Cp(Y ) is nonmetrizable and Fréchet–Urysohn. 310 15 Sequential Properties in the Class G

AtvsE is said to have a superresolution if E admits a bounded resolution N {Kα : α ∈ N } such that for every ﬁnite tuple (n1,...np) of positive integers and ev- N ery bounded set Q in E there exists α = (mk) ∈ N such that mj = nj for 1 ≤ j ≤ p and Kα absorbs Q. This implies that for any ﬁnite tuple (n1,...np) the sequence

~~(Cn1,...,np,n)n is bornivorous in E. Clearly, every metrizable tvs admits a superresolution. We note the following proposition.~~

~~Proposition 15.1 A b-Baire-like space E is metrizable if and only if E admits a superresolution.~~

N Proof Let {Kα : α ∈ N } be a superresolution for E consisting of absolutely convex bounded sets. Then the sets Cn1n2,...,nk are also absolutely convex. Note that for N every α = (nk) ∈ N and every neighborhood of zero U in E there exists k ∈ N ⊂ k such that Cn1,n2,...,nk 2 U by applying the proof of Proposition 7.1.IfF is the N completion of E, the space F is Baire-like. We prove that there exists α = (nk) ∈ N ∈ N such that Cn1,n2,...,nk is a neighborhood of zero in F for each k . Since, by the assumption, the sequence (nCn)n is bornivorous in E, and E is quasibarrelled, we apply Proposition 2.13 to deduce that F = E = nCn ⊂ (1 + ε) nCn n n

~~∈ N for each ε>0. Since F is a Baire-like space, there exists n1 such that Cn1 is a neighborhood of zero in F .~~

Now assume that for a ﬁnite tuple (n1,...np) of positive integers the set Cn1,...,nk ≤ ≤ is a neighborhood of zero for each 1 k p. By the assumption, the sequence = (nCn1,...,np,n)n is bornivorous in E, and consequently E n nCn1,...,np,n. We ap- ∈ N ply the same argument as above to get np+1 such that Cn1,...,np,np+1 is a neighborhood of zero. This completes the inductive step. We provided a countable basis −k (2 Cn1,...,nk )k of neighborhoods of zero, so E is metrizable.

~~Corollary 15.7 Cp(X) is a metrizable space if and only if Cp(X) admits a superresolution.~~

Proof First observe that for any X the space Cp(X) is b-Baire-like. Indeed, let (An)n be a bornivorous sequence of absolutely convex closed subsets of Cp(X) covering Cp(X). Cp(X) is quasibarrelled [213, Corollary 11.7.3]. We apply Propo- RX = = RX sition 2.13 to deduce Cp(X) n An, where the closure is taken in .By X the Baire category theorem, some Am is a neighborhood of zero in R . Hence Am is a neighborhood of zero in Cp(X),soCp(X) is b-Baire-like. Now Proposition 15.1 completes the proof. 15.2 Sequential (LM)-spaces and the dual metric spaces 311

~~15.2 Sequential (LM)-spaces and the dual metric spaces~~

In the previous section, we showed that any Fréchet–Urysohn lcs in the class G is metrizable. One can ask if the same conclusion holds for any sequential lcs E;for the deﬁnition, see the text before Corollary 9.15. Clearly, every Fréchet–Urysohn space is sequential, and the converse implication fails in general. Nyikos [316] showed that the ℵ0-dimensional space ϕ is sequential and is not Fréchet–Urysohn. Since ϕ is the strong dual of the Fréchet space ω := KN of all scalar sequences and is an (LB)-space, it is a Montel (DF )-space. It should be pointed out that Webb [415] used the designations C1 and C2 for the Fréchet–Urysohn and sequential spaces, respectively. According to Webb [415], a topological space E is said to have the property C3 if the sequential closure of any subset of E is sequentially closed. Clearly,

~~metrizable ⇒ C1 ⇔[C2 ∧ C3].~~

The property C3 has been used in [224] to show that an (LM)-space E is metrizable if and only if it has the property C3. Nevertheless, in [229] it was proved that there exist nonmetrizable (DF )-spaces with the property C3. This section and the next are based on results from [229].

~~Proposition 15.2 Any sequential topological space E has countable tightness.~~

~~Proof Assume E is a sequential space. Fix an arbitrary set M ⊂ E. For any x ∈ M, we need to obtain a countable subset, say Mx ⊂ M, such that x ∈ Mx . Set N := {A : A ⊂ M, and |A|≤ℵ0}.~~

Clearly, N ⊂ M. Observe that N is a sequentially closed set. Take {xn,n∈ N}⊂N such that xn → y ∈ E. For every n ∈ N, there exists a countable set An ⊂ M such ∈ ∈ that xn An. Therefore y n An and, by the deﬁnition of N, we conclude that y ∈ N. Since E is sequential, the set N is closed, and so x ∈ N = M.

The converse implication fails in general. Corollary 9.15 provides a large class of topological groups X such that (X, σ (X, X∧)) has countable tightness and X is not sequential. Yoshinaga [436] proved that every Silva space (equivalently, the strong dual of a Fréchet–Schwartz space) is sequential. Webb [415] extended this result to all Montel (DF )-spaces (equivalently, strong duals of Fréchet–Montel spaces) and also proved that only ﬁnite-dimensional Fréchet–Montel spaces are Fréchet– Urysohn. We know already that proper (LB)-spaces are not Fréchet–Urysohn spaces (see Corollary 15.5). In particular, the nonmetrizable space ϕ (which is a proper (LB)- space) provides an example of a space in the class G that is not Fréchet–Urysohn. Nyikos asked if the direct sum of countably many (complete) metrizable groups endowed with the box product topology is sequential. Theorem 15.2 and Theorem 15.3 provide examples of the Nyikos question. 312 15 Sequential Properties in the Class G

Recall again that an lcs E is ∞-quasibarrelled if every β(E,E)-bounded sequence in E is equicontinuous, and it is a dual metric space if it is ∞-quasibarrelled and admits a fundamental sequence of bounded sets. We will need a couple of lemmas.

Lemma 15.3 Let E be either a dual metric or an (LM)-space. The following assertions are equivalent for E: (i) Every bounded set in E is relatively sequentially compact. (ii) Every bounded set in E is relatively countably compact. (iii) Every bounded set in E is relatively compact (i.e., E is semi-Montel). (iv) E is a Montel space (i.e., E is barrelled and semi-Montel).

Proof By Theorem 10.2 and Theorem 10.3, the space E is angelic. Now the conditions (i), (ii) and (iii) are equivalent by the well-known fact stating that in angelic spaces (relatively) sequential compact = (relatively) countable compact = (relatively) compact; see [165]. Clearly, (iv) ⇒ (iii). (iii) ⇒ (iv): If (iii) holds, E is quasicomplete (meaning that every bounded closed set in E is complete) and hence sequentially complete. If E is an (LM)-space, then (as it is quasibarrelled) E is barrelled; see also [246, Proposition 27.1 (1)]. Hence (iv) holds for (LM)-spaces. If E is a dual-metric space, every bounded subset of E is precompact (by the assumption), and we apply Theorem 10.3 to deduce that every bounded set in E is metrizable. Hence E is separable (being the countable union of metrizable bounded subsets). As every separable ∞-quasibarrelled space is quasibarrelled [328, Theorem 8.2.20], the sequentially complete space E is barrelled. Hence E is a Montel space.

We know that every sequential topological space has countable tightness by Proposition 15.2. Note also that every sequential ∞-quasibarrelled space E is a Mackey space (i.e., the original topology of E equals the Mackey topology of E).

~~Proposition 15.3 If E is an ∞-quasibarrelled space with countable tightness, it is a Mackey space. Hence every sequential ∞-quasibarrelled space is a Mackey space.~~

Proof Let ξ be the original topology of E. Assume that ξ is not the Mackey topology μ := μ(E, E). Then there exists a μ-closed set B that is not ξ-closed. Fix x ∈ B \ B, where the closure is taken in ξ. Choose a μ-continuous seminorm such that p(x − y) ≥ 1 for all y ∈ B. Fix an arbitrary countable set {xn : n ∈ N} in B. Using the Hahn–Banach theorem, we select a sequence (fn)n in E such that

fn(x − xn) = 1, |fn(z)|≤p(z), for all n ∈ N and all z ∈ E. This implies that the sequence (fn)n is μ-equicontinuous. Hence there exists an absolutely convex σ(E,E)-compact subset containing the elements of the sequence (fn)n. Thus (fn)n is β(E ,E)-bounded. Since (E, ξ) is 15.2 Sequential (LM)-spaces and the dual metric spaces 313

~~∞ -quasibarrelled, the sequence (fn)n is ξ-equicontinuous. Hence its polar set U is a ξ-neighborhood of zero. Consequently, the set~~

~~− V := x + 2 1U is a ξ-neighborhood of x but xn ∈/ V for all n ∈ N. This shows that (E, ξ) does not have countable tightness, a contradiction.~~

~~Lemma 15.4 provides a much stronger variant of Proposition 15.3 and extends Webb’s result [415, Theorem 5.5(i)]; see [229].~~

~~Lemma 15.4 If E is a sequential ∞-quasibarrelled lcs, the space E is either b-Baire-like or barrelled in which every bounded set is relatively sequentially compact.~~

Proof First note that E is quasibarrelled. Indeed, let ξ be the original topology of E. Let ξ0 be the locally convex topology on E of the uniform convergence on β(E ,E)- bounded sets. Clearly, ξ ≤ ξ0 and ξ0 is quasibarrelled. We show that ξ = ξ0 by proving that every ξ0-closed set is ξ-closed. Note that every ξ-null sequence is a ξ0-null sequence. Assume that there exists a sequence (xn)n in E such that xn → 0 in ξ but xn 0inξ0. This means that there exist a β(E ,E)-bounded set B ⊂ E ,

~~ε>0, and a subsequence (xnk )k of (xn)n, as well as a sequence (fk)k in B such that | | ∈ N fk(xnk ) >εfor all k . We know that~~

~~−1 −1 ε {fk : k ∈ N}⊂ε B~~

~~ ◦ is a β(E ,E)-bounded set, so the polar ε{fk : k ∈ N} is a neighborhood of zero in ξ (since ξ is ∞-quasibarrelled) such that~~

∈ { : ∈ N}◦ xnk / ε fk k for each k ∈ N. Hence we obtain the contradiction xn 0inξ. Since E is sequential and both topologies ξ and ξ0 have the same convergent sequences, we have that each ξ-closed subset of E is also ξ0-closed. Hence ξ = ξ0 as claimed. Now assume that E is not b-Baire-like. We need to show that E is barrelled and every bounded set in E is relatively sequentially compact. Assume by way of contradiction that E contains a bounded sequence (zn)n that does not have a convergent subsequence. Since E is quasibarrelled and not b-Baire-like, there exists an increasing bornivorous sequence (Bn)n of closed, absolutely convex sets covering E such that for each set Bn there exists a bounded set not absorbed by Bn. Taking a sub- −1 sequence, if necessary, we may assume that n zk ∈ Bn for each k ∈ N. For each 2 −1 n ∈ N, there exists a bounded sequence (vk,n)k that misses n Bn.Ifuk,n := n vk,n, then for the null sequence (uk,n)k we have

~~−1 uk,n ∈/ nBn,uk,n = −n zk, 314 15 Sequential Properties in the Class G for each k ∈ N. Hence~~

~~−1 0 ∈/ An := {n zk + uk,n : k ∈ N}. := We reach a contradiction by showing that the set A n An is sequentially closed but not closed.~~

~~Claim 15.1 The set A is not closed in E.~~

Indeed, take a neighborhood of zero U in E. Then there exists n ∈ N such that −1 n zk ⊂ U for all k ∈ N. There exists k0 ∈ N such that uk,n ∈ U for all k ≥ k0. Finally, −1 n zk + uk,n ∈ A ∩ (U + U)= ∅ for all k ≥ k0. This shows that 0 ∈ A \ A. The claim is proved.

~~Claim 15.2 The set A is sequentially closed.~~

In fact, let (xj )j be a sequence in A converging to x ∈ E. Note that only finitely many distinct elements of xj are included in any An. Indeed, otherwise we can −1 find a subsequence of the sequence (n zk + uk,n)k that would converge to ele- −1 ment x and then the corresponding subsequence of (n zk)k would converge to x − limk uk,n = x. This provides a contradiction since then a subsequence of (zk)k would converge. Something more is even true: Only finitely many An contain one element of the sequence (xj )j . Indeed, assume that the sequence (xj )j has elements −1 of the form n zn + uk,n for arbitrary large n ∈ N. Then, since the sequence (uk,n)k is unbounded (because un,k ∈/ nBn for k,n ∈ N and (Bn)n is bornivorous) and (zk)k is bounded, the corresponding sequence (xj )j of sums is an unbounded converging sequence, a contradiction. This implies that only finitely many of the sets An may contain elements of the sequence (xj )j and, as we proved above, each such set An contains only finitely many distinct elements of (xj )j . This yields the conclusion that the sequence (xj )j has only finitely many distinct elements. Therefore there exists m ∈ N such that x = limk xj = xm ∈ A. We proved the claim. Since we proved that, if E is not b-Baire-like, the space E is sequentially complete, we deduce that the quasibarrelled space E is sequentially complete. Hence E is barrelled.

~~Now we are ready to prove the following theorem.~~

~~Theorem 15.2 (Kakol–Saxon) ˛ The following conditions are equivalent for a dual metric space E: (i) E is sequential. (ii) E is normable or a Montel (DF )-space.~~

~~Proof By (Sn)n we denote a fundamental sequence of bounded sets in E consisting of closed absolutely convex subsets of E. 15.2 Sequential (LM)-spaces and the dual metric spaces 315~~

(i) ⇒ (ii): If E is nonnormable, it does not admit a bounded neighborhood of zero. Then E cannot be b-Baire-like. By Lemma 15.4, it follows that E is a barrelled space for which each bounded set is relatively sequentially compact. Now Lemma 15.3 is applied to deduce that E is a Montel (DF )-space. (ii) ⇒ (i): Assume E is a Montel space. It sufﬁces to show that if A ⊂ E is a sequentially closed set such that 0 ∈/ A, then 0 ∈/ A.Let(Sn)n be a fundamental sequence of bounded, absolutely convex subsets of E. There exists ε1 > 0 such that

A ∩ ε1S1 =∅ since A does not contain sequences converging to zero. Since ε1S1 is sequentially compact by Lemma 15.3,thesetA − ε1S1 is sequentially closed and does not contain zero. Hence there exists ε2 > 0 such that (A − ε1S1) ∩ ε2S2 =∅, and the compact set ε1S1 + ε2S2 is sequentially compact. Continuing this procedure, we obtain a sequence (εn)n of positive numbers such that Kn∩ A =∅for each n ∈ N, where := := Kn k≤n εkSk. On the other hand, the set K n Kn is absolutely convex and absorbs bounded sets of E. As any Montel (DF )-space is an (LB)-space, the space E is bornological. Hence K is a neighborhood of zero. As K ∩ A =∅,wehave 0 ∈/ A.

~~For (LM)-spaces, we note the following theorem.~~

~~Theorem 15.3 (Kakol–Saxon) ˛ The following assertions are equivalent for an (LM)-space E: (i) E is sequential. (ii) E is metrizable or a Montel (DF )-space.~~

Proof (ii) ⇒ (i): Apply (ii) ⇒ (i) in the proof of Theorem 15.2. (i) ⇒ (ii): Since E is an (LM)-space, it is bornological and consequently ∞-quasibarrelled. Assume E is nonmetrizable. We prove that it is a Montel (DF )-space. Exactly as it was proved in Saxon and Narayanaswami’s paper [356] for (LF )-spaces, one obtains that E is not b-Baire-like. Then Lemma 15.4 and Lemma 15.3 imply that E is sequentially complete and Montel, so it is an (LF )- space. Now it is enough to show that E admits a fundamental sequence of bounded sets, since then, being barrelled, it will be a (DF )-space. By Corollary 2.7, we deduce that E contains an isomorphic copy of ϕ.Letξ be the original topology of E. Let (En,ξn)n be a deﬁning sequence for E of Fréchet spaces. We claim that for each n ∈ N there exists an absolutely convex neighborhood of zero Wn in (En,ξn) that is bounded in E. Assume the claim fails. Then there exists (Em,ξm) such that no ξm-neighborhood of zero is a bounded set in E.Let(Un)n be a decreasing basis of absolutely convex neighborhoods of zero in the space (Em,ξm). As each set Un is unbounded, for each n ∈ N there exists fn ∈ E such that fn(Un) = K, where K denotes the scalar ﬁeld of E.By(en)n, we denote a Hamel basis for the space ϕ ⊂ E. For

~~y = a1e1 +···+akek ∈ ϕ 316 15 Sequential Properties in the Class G with ak = 0, set~~

~~l(y) = k, v(y) = ak, y:=sup {|a1|, |a2|,...,|ak|}. Set~~

−1 A := {x + y : 0 = y ∈ ϕ,y≤1,x∈ U l(y),fl(y)(x + y) =[v(y)] }, where the closure is taken in E. We show that A is sequentially closed and not closed in E. This contradiction will prove the claim. A is not closed in E. Indeed, clearly 0 ∈/ A since each member of A has a nonzero image under a suitable linear functional. We show that 0 ∈ A \ A. Let U and V be neighborhoods of zero in E such that V + V ⊂ U. There exists k ∈ N such that Uk ⊂ V . It is easy to select a nonzero element y ∈ V ∩ ϕ such that l(y) = k and −1 y≤1. Now choose a point x ∈ Uk such that fk(x) =[v(y)] − fk(y). Then x + y ∈ A ∩ (V + V)⊂ A ∩ U.

This means that 0 ∈ A \ A. A is sequentially closed in E. Indeed, let (zn)n be a sequence in A that converges to z ∈ E. Then, applying the definition of A, we note that there exists nonzero yn ∈ ϕ such that ≤ := − ∈ =[ ]−1 yn 1,xn zn yn U l(yn),fl(yn)(zn) v(yn) , for all n ∈ N. Note that m := sup{l(yn) : n ∈ N} is finite. Indeed, if this is not true, there exists a subsequence l(ynk )k of l(yn)n tending to infinity. Since for every ∈ N ⊂ closed neighborhood of zero U in E there exists m such that U l(yk) U for all ≥ → → k m,wehavexnk 0. Hence ynk z, so the sequence (ynk )k defines an infinite- dimensional bounded subset in the space ϕ. This is impossible, as all bounded sets in ϕ are finite-dimensional. Set

L := span{e1,e2,...,em}, and let γ be a topology generated by the norm y. Then ξ|L = γ |L. This shows that the sequence (yn)n is ξ-bounded in L. The classical Bolzano–Weierstrass theorem [213] is applied to select a convergent subsequence → ∈ (ymk )k y ϕ ≤ = of a subsequence of (yn)n such that there exists p m with l(ymk ) p for each k ∈ N. Then y≤1 and

~~−1 fp(z) = lim fp(zm ) = lim [v(ym )] ; k k k k therefore limk v(ymk )>0 and then the preceding equalities and the continuity of each coefﬁcient functional imply that~~

~~−1 −1 fp(z) = lim v(ym ) =[v(y)] k k 15.2 Sequential (LM)-spaces and the dual metric spaces 317~~

~~= ⊂ and p l(y). Finally, since xmk U l(y), we note that~~

x = z − y = lim (zm − ym ) ∈ U l(y). k k k Hence z = x + y ∈ A, which proves that A is sequentially closed. This contradiction with assumption (i) shows that the claim holds. Using the claim, we obtain an increasing sequence (Sn)n of absolutely convex bounded sets in ξ covering E such that each Sn is a ξn-closed neighborhood of zero in ξn. Note that (Sn)n is a fundamental sequence of ξ-bounded sets, where the closure is taken in ξ. Indeed, assume that there exists a bounded set B in (E, ξ) that is not −1 absorbed by any Sn. Then, for each n ∈ N there exists xn ∈ n B such that xn ∈/ Sn. For each n ∈ N,letVn be a ξn-neighborhood of zero such that xn ∈/ Sn +Vn. Clearly, the sequence (Vn)n can be taken as decreasing. Then Sk ∩ Vk ⊂ Sn + Vn for each n, k ∈ N. Consequently, V ⊂ Sn + Vn, where V =: ac (Sk ∩ Vk) k is a ξ-neighborhood of zero in E. This proves xn ∈/ V for each n ∈ N.Weshowed that the neighborhood of zero V in ξ misses all elements of the null sequence (xn)n, a contradiction. The proof is completed.

~~Theorem 15.3 is applied to provide an example of a Montel (LF )-space that is not a (DF )-space.~~

~~Example 15.1 The space E := RN × ϕ is a Montel (LF )-space that is not a (DF )- space; hence E is not sequential.~~

~~Proof Note that RN is not normable, so (by the Baire category theorem) does not have a fundamental sequence of bounded sets. Theorem 15.3 shows that E is not sequential.~~

Example 15.1 provides another approach to produce examples of a nonsequential lcs with countable tightness (recall that any (LF )-space has countable tightness; see Theorem 12.3). We complete this section with the following converse to Webb’s result [415, The- orem 5.5 (1)].

~~Proposition 15.4 (Kakol–Saxon) ˛ Every sequential proper (LB)-space is a Montel space.~~

Proof Since E is a proper (LB)-space, it contains ϕ; see Corollary 2.6.Asevery bounded set in ϕ is ﬁnite-dimensional, we can select in E a linearly independent 318 15 Sequential Properties in the Class G sequence (yn)n whose linear span does not admit an inﬁnite-dimensional bounded set. By Lemma 15.3, it is enough to show that every bounded sequence (zn)n in E contains a convergent subsequence. Assume this is not true, and let (zn)n be a bounded sequence that does not contain a convergent subsequence. Set

~~−1 −1 A := {n zk + k yn : k,n ∈ N}.~~

The set A is not closed since 0 ∈ A \ A. Indeed, choose an arbitrary neighborhood of zero V in E.LetU be a neighborhood of zero in E such that U + U ⊂ V .The −1 sequence (zk)k is bounded, so there exists n ∈ N such that n zk ⊂ U for each −1 k ∈ N. For this n ∈ N, choose k ∈ N such that k yn ⊂ U. Then there exist k,n ∈ N such that −1 −1 n zk + k yn ∈ U + U ⊂ V. This proves that 0 ∈ A \ A. Now we prove that A is sequentially closed by showing that every convergent sequence (n−1z + k−1y ) in A has only finitely many p kp p np p { } { −1 } distinct members. If the set p np were infinite, the set p kp ynp would be an { } infinite-dimensional bounded set, a contradiction. Then the set p np is finite and, without loss of generality, we may assume that np = m for some fixed m ∈ N. Then, { } −1 = −1 if p kp were infinite, limp kp ym 0, implying that the sequence (m zkp )p, and then (zkp )p also, would be convergent, which provides another contradiction. This argument proves that A is sequentially closed and not closed. Hence E is not sequentially closed, giving a contradiction. We proved that E is Montel.

AtvsE is said to have the property C4 if for each sequence (xn)n in E there exists a sequence (tn)n of positive scalars such that 0 ∈ {tnxn : n ∈ N}. The following lemma motivates the next proposition.

~~Lemma 15.5 For a tvs E, the property C3 implies C4. There exists a set Γ such that the space ∞(Γ ) endowed with the topology of pointwise convergence has the property C4 but not C3.~~

−1 Proof Let (xn)n be a sequence in E. Choose a nonzero vector a ∈ E \{k nxn : k,n ∈ N}. Set −1 −1 H := {n a − k xn : k,n ∈ N}, and denote by H − the sequential closure of H . Then n−1a ∈ H − for all n ∈ N, and 0 belongs to the sequential closure H −− of H −.IfE isassumedtohavethe − property C3,wehave0∈ H . Therefore there are sequences (np)p and (kp)p in N such that −1 − −1 → →∞ np a kp xnp 0,p .

Note that (np)p is unbounded. Indeed, otherwise, taking a subsequence, we may ∈ N −1 − −1 → assume that there exists m such that m a kp xm 0, and if (kp)p is unbounded, we have a = 0, and if (kp)p is bounded, there exists r ∈ N such that 15.2 Sequential (LM)-spaces and the dual metric spaces 319

~~−1 m a − rxm = 0. These two cases provide a contradiction. We may assume that −1 → (np)p is strictly increasing. Clearly, (kp xnp ) 0. Deﬁne~~

~~:= −1 ∈ N = ∈ N \{ : ∈ N} tnp kp ,p ,tn 1,n np p .~~

Then the sequence (tn)n is as required. Now we prove the other part of the lemma. Let Σ be the set of sequences of different pairs (i, j) ∈ N × N such that for every σ ∈ Σ and every j ∈ N there exists at most one i ∈ N with (i, j) ∈ σ . Set Γ = Σ ∪ N, and let ∞(Γ ) be the space of bounded functions on Γ with the topology of = γ the pointwise convergence. Set x(i,j) (x(i,j))γ ∈Γ , where xγ = 0, if γ ∈ Σ, (i,j) ∈ γ or if γ ∈ N,j= γ, xγ = (i,j) (i,j) γ = ∈ ∈ ∈ N = x(i,j) 1, if γ Σ, (i,j) γ, or if γ ,j γ.

~~∈ N → →∞ := γ Then, for each j , one has x(i,j) ej , i , where ej (ej )γ ∈Γ is deﬁned by 1ifγ ∈ N,j= γ, eγ = j 0 otherwise.~~

~~Also, ej → 0ifj →∞. Set~~

~~B := {x(i,j) : (i, j) ∈ N × N}.~~

Then zero does not belong to the sequential closure of B. Indeed, otherwise we can find in B a sequence x → 0(i.e.,xγ → 0 for each γ ∈ Γ ). But then (in,jn) (in,jn) either there exists j0 ∈ N such that the set A := {(in,jn)} contains infinite, many pairs (i, j0) or there exists in A a subsequence (ink ,jnk )k in Σ. The first case for γ := j and the other one for γ := (i ,j ) yields a sequence xγ that does not 0 nk nk k (in,jn) converge to zero since it contains a constant subsequence convergent to 1, a contradiction. ∞ We showed that (Γ ) does not have the property C3, although it satisfies the property C4.

~~Proposition 15.5 (Kakol–Saxon) ˛ For a (DF )-space E, the following assertions are equivalent: (i) E does not contain ϕ. (ii) E admits a ﬁner normed topology. (iii) E has the property C4.~~

Proof (i) ⇒ (ii): Let (Sn)n be a fundamental sequence of absolutely convex bounded closed sets. It is enough to prove that there exists m ∈ N such that Sm is absorbing in E. Then the Minkowski functional norm

−1 x:=inf {t>0 : t x ∈ Sm} 320 15 Sequential Properties in the Class G generated by the set Sm will provide a norm as required. By a contradiction, assume that none of Sn is absorbing in E. Taking a subsequence of (Sn)n if necessary, we select a sequence (xn)n such that

~~xn ∈ Sn+1 \ span (Sn), n ∈ N. (15.1)~~

Set S to be the linear hull of {xn : n ∈ N}. Clearly, xn,n∈ N, are linearly independent. We prove that the induced topology on S is the ﬁnest locally convex topology. Let p be a seminorm on S. It is enough to show that p is continuous. Since (15.1) holds, we follow the proof of Theorem 2.5 (see also [354, Lemma]) to get a sequence ∈ ∩ ◦ ∈ N fn E Sn, n , such that −n max |fr (x)|≥(1 + 2 )p(x) r≤n

~~= := | | for x 1≤i≤n aixi . Set q(x) supn fn(x) . Then {x ∈ E : q(x) ≤ 1}∩Sn = {x ∈ E :|fr (x)|≤1}∩Sn r = {x ∈ E :|fr (x)|≤1}∩Sn. 1≤r~~

This holds since |fr (x)|≤1forx ∈ Sn and r ≥ n. We showed that {x ∈E : q(x)≤1} ∩ Sn is a neighborhood of zero in Sn for each n ∈ N. Since E is a (DF )-space, {x ∈ E : q(x) ≤ 1} is a neighborhood of zero in E (see [246, 29.3(2)], or [328, Corollary 8.3.3]). Consequently, q is continuous on E, and then p is continuous on S. We proved that E contains ϕ, which provides a contradiction with (i). (ii) ⇒ (i)isobvious. (ii) ⇒ (iii) is clear since a normed topology has the property C4. (iii) ⇒ (ii): Again it is enough to show that there exists m ∈ N such that Sm is absorbing. Assume this is not true. Then, for each n ∈ N there exists xn in E that is not absorbed by Sn. By (iii), there exists a sequence (tn)n of positive scalars such that zero is in the closure {tnxn : n ∈ N}. Clearly, tnxn ∈/ Sn for all n ∈ N. Since each set Sn is closed, for each n ∈ N there exists an absolutely convex neighborhood of zero Un in E such that

tnxn ∈/ Sn + Un. := + Set U n Sn Un. It is known that U is a neighborhood of zero in E;see[328, Proposition 8.3.5]. This yields a contradiction since then tmxm ∈ U for some m ∈ N, and the proof is completed.

~~− 15.3 (LF )-spaces with the property C3~~

First observe that in the class of quasibarrelled (DF )-spaces E the property C3 implies the metrizability of E;see[229] and the proof of [246, Theorem 29.1]. − 15.3 (LF )-spaces with the property C3 321

~~Proposition 15.6 A quasibarrelled (DF )-space E has the property C3 if and only if E is normable.~~

Proof Let (Sn)n be a fundamental (increasing) sequence of closed, absolutely convex bounded sets in E. Assume E is not normable. Since E is quasibarrelled, for each Sn there exists in E a bounded absolutely convex set An such that Sn does not absorb An. Hence, for each pair (n, k) ∈ N × N, there exists

~~−1 zn,k ∈ k An \ (k + 1)Sn.~~

~~Choose a sequence (xn)n of nonzero elements of A1 such that xn → 0. Clearly, ⊂ ∈ N → →∞ ∈ N A1 Sm1 for some m1 and zn,k 0ifk , n . Set~~

~~M := {xn + zn,k : n, k ∈ N}.~~

− Note that each xn belongs to the sequential closure M of the set M. On the other − hand, 0 ∈/ M . Indeed, otherwise there exist sequences (np)p and (kp)p in N such that np →∞and + → xnp znp,kp 0.

~~Hence there exists r>m1 such that + ∈ xnp znp,kp Sr for all np,kp ∈ N. This implies that ∈ + ⊂ + = znp,kp Sr Sm1 Sr Sr 2Sr .~~

~~Hence np ≤ r, a contradiction. The converse implication is obvious.~~

This shows that the space ϕ contains a subset whose sequential closure is not sequentially closed in ϕ. Bonet and Defant [61] proved that if E is an infinite-dimensional nuclear (DF )- space different from ϕ, the space E contains a subspace whose sequential closure is not a sequentially closed set. This motivates us to distinguish a class of tvs having − the following property: A tvs E is said to have the property C3 if the sequential closure of any linear subspace of E is sequentially closed; see [229]. We will see that the only infinite-dimensional Montel (DF )-space with the prop- − − erty C3 is the space ϕ.An(LF )-space E has the property C3 if and only if E is isomorphic to some metrizable (LF )-space M,toϕ, or to the product space M × ϕ; see [229, Theorem 6.13]. A corresponding characterization for (LB)-spaces with − the property C3 is also provided in [229, Corollary 6.12] and will be presented in Theorem 15.5. AtvsE is said to be docile if every infinite-dimensional subspace of E admits an infinite-dimensional bounded subset. The next proposition uses a deep result of Josefson an Nissenzweig [215], [313]. 322 15 Sequential Properties in the Class G

~~Theorem 15.4 A Banach space E is ﬁnite-dimensional if and only if every sequence in E which σ(E,E)converges to zero converges to zero in the norm topology of the dual E.~~

A natural extension of this result to Fréchet spaces is due to Bonet, Lindström, Schlumprecht and Valdivia; see [63] for details. A simple relation between the docility and the property C3 shows the following.

~~∗ Proposition 15.7 Every tvs with the property C3 is docile. The weak dual of any inﬁnite-dimensional Banach space E is docile without the property C3.~~

Proof Assume E is not docile. Then there exists a linearly independent sequence (yn)n in E whose linear span does not contain any inﬁnite-dimensional bounded set. Set −1 −1 A := {n y1 + k yn : k,n ∈ N}. − −1 Then the sequential closure A of A equals the set A ∪{n y1 : n ∈ N}. Since − − 0 ∈/ A , A is not sequentially closed. Therefore E does not have the property C3. Now assume E is an inﬁnite-dimensional Banach space. Clearly, the space (E,σ(E,E)) is docile. Let B be the closed unit ball in E. By Theorem 15.4, it follows that B contains a linearly independent sequence (fn)n that converges to zero in σ(E,E). Now we proceed as before. Set

~~−1 S := {fn + n fm : n, m ∈ N}. − − Then the sequential closure S equals the set S ∪{fn : n ∈ N}. Since 0 ∈/ S ,the sequential closure of S is not sequentially closed.~~

~~To prove the main result of this section, we need a few additional lemmas.~~

~~Lemma 15.6 Let E be an inﬁnite-dimensional docile tvs. Then E contains a sequentially dense subspace F such that dim(E) = dim(E/F ).~~

Proof Case 1. Assume dim(E) =ℵ0. Then there exists a subspace G ⊂ E such that dim(G) = dim(E/G) =ℵ0. By the assumption, there exists in G an inﬁnite- dimensional bounded set B. Hence we can choose a double indexed set

~~S := {zn,k : k,n ∈ N}⊂B of linearly independent elements. Choose a Hamel basis {xn : n ∈ N} for an algebraic complement of span(S) in E. Set~~

~~−1 F := span{xn + k zn,k : n, k ∈ N}.~~

− −1 Then xn ∈ F (the sequential closure) for each n ∈ N since xn + k zn,k → xn if k →∞for each n ∈ N. Then − zn,k = (zn,k + xn) − xn ∈ F , − 15.3 (LF )-spaces with the property C3 323 and hence E ⊂ F −.ThisprovesCase1. Case 2. dim(E) =ℵ0 · dim(E). Then there exists a family {Eα : α ∈ A} of vector | |= =ℵ ∈ = subspaces of E such that A dim(E), dim(Eα) 0 for each α A, and E α Eα. The ﬁrst case provides a family of sequentially dense proper subspaces ⊂ := =| |= Fα Eα. Then F α Fα is sequentially dense in E and dim(E/F ) A dim(E). The proof is completed.

Lemma 15.7 Let E be an lcs with an increasing bornivorous sequence (Sn)n of subsets of E. Assume that E contains docile inﬁnite-dimensional subspaces Gn ⊂ span(Sn) such that Gn+1 ∩ span(Sn) ={0}, n ∈ N. Then E does not have − the property C3 .

Proof Applying Lemma 15.6, we obtain a sequentially dense subspace F ⊂ G1 and x ∈ G1 \ F .Let(yn)n be a sequence in F with yn → x. We may assume that the set {x}∪{yn : n ∈ N} is linearly independent, passing to a subsequence if necessary. We claim that there exist sequences (fn)n ⊂ E and (zr,s )s ⊂ E such that: (i) fn(x) = 0 for all n ∈ N. (ii) fn(yi) = 0 for all i

~~fk+1(x) = 0,fk+1(T \{x}) ={0}.~~

~~Next, choose linearly independent~~

~~∈ ∩ ⊥ ∩ ⊥ ∩···∩ ⊥ zk+1,1,zk+1,2,...,zk+1,k+1 Gk+2 f1 f2 fk+1, and for each n ≤ k we select~~

~~∈[ ∩ ⊥ ∩ ⊥ ∩···∩ ⊥ ]\ { : ≤ } zn,k+1 Gn+1 f1 f2 fk+1 span zn,s s k .~~

~~The claim is proved. We may assume that each sequence (zn,s)s is bounded since the space span{zn,s : s ∈ N} contains an inﬁnite-dimensional bounded set (by the docility). Set~~

~~−1 H := span{yn + s zn,s : n, s ∈ N}.~~

~~− −1 Then, as before, yn ∈ H for each n ∈ N since yn + s zn,s → yn for s →∞. −− Hence x ∈ H since yn → x if n →∞. 324 15 Sequential Properties in the Class G~~

∈ − − We show that x/H , and this will show that E does not have the property C3 . Assume that there exists a sequence (xn)n ⊂ H such that xn → x. Then there exists m ∈ N such that Sm absorbs all the set {xn : n ∈ N}. Note that

~~−1 span(Sm) ∩ H = span{yn + s zn,s : n~~

~~This implies that fm(xn) = 0 for all n ∈ N. From (i) it follows that fm(x) = 0, so the conclusion holds.~~

~~Now we are ready to prove the following main result; see [229].~~

~~Proposition 15.8 (Kakol–Saxon) ˛ The only inﬁnite-dimensional Montel (DF )- − space E with the property C3 is the space ϕ.~~

Proof Let (Sn)n be a fundamental (increasing) sequence of closed, bounded, absolutely convex sets in E. Since E is Montel, we may assume that Sn’s are compact. Case 1. For each n ∈ N, the space span(Sn) is infinite-codimensional in span(Sn+1). We may assume that G1 := span(S1) is infinite-dimensional. By induction, we obtain a sequence (Gn)n of subspaces of E such that Gn+1 ⊂ span(Sn+1) is infinite-codimensional for each n ∈ N and Gn+1 ∩ span(Sn) ={0}. Since each Gn admits a stronger normed topology generated by the Minkowski functional norm associated with the set Sn, each Gn is docile. By Lemma 15.7, we conclude that E − does not have the property C3 . Case 2. There exists m ∈ N such that span(Sm) is finite-codimensional in the space span(Sm+1). Since (Sn)n is increasing, we may assume that each span(Sn) has finite codimension in span(Sn+1). Hence Hn := span(Sn) is countable- codimensional in E for each n ∈ N. Since every countable-codimensional subspace of a barrelled space is barrelled [328], each Hn is a barrelled subspace of E having a compact neighborhood of zero Sn. Therefore each Hn is finite-dimensional, consequently yielding the countable-dimensionality of the whole space E,soE is a barrelled space of dimension ℵ0. Hence E is isomorphic to the space ϕ by applying in Proposition 2.14.

~~Proposition 15.9 Let M be a metrizable lcs. Then the product M × ϕ has the − property C3 and does not have the property C3.~~

~~Proof Let {en : n ∈ N} be a Hamel basis in ϕ. Set~~

Sn := {(x, y) : x ∈ M,y ∈ n ac{e1,e2,...,en}} for each n ∈ N. Note that each Sn is a closed subset of M × ϕ, (Sn)n covers the whole space M × ϕ, and (Sn)n is increasing bornivorous in M × ϕ. Let Fn := span(Sn) for each n ∈ N. Clearly, each Fn is metrizable and has codimension 1 in the space Fn+1, and Sm ∩ Fn is closed for each m>n. We need to := × − show that E M ϕ has the property C3 : Fix a linear subspace H of E.We − = −− − show that H H , which will show that E has the property C3 . Fix arbitrary − 15.3 (LF )-spaces with the property C3 325

−− − x ∈ H . Then there exist sequences (yn)n ⊂ H with yn → x, (zn,k)k ⊂ H with zn,k → yn for each n ∈ N. There exists p ∈ N such that {yn : n ∈ N}⊂pSp (since (Sn)n is increasing bornivorous), and for each n ∈ N there is mn >pwith { : ∈ N}⊂ zn,k k mnSmn .

~~Note that there exists a sequence (wn,k)k such that~~

wn,k ∈ 2ac{zn,k : k ∈ N}∩Fp and such that wn,k → yn for each n ∈ N. Indeed, fix n ∈ N. Note that Fp is finite- ∩ = ⊕ codimensional in Fmn and Smn Fp is closed. Let Fmn Fp Lp algebraically, where Lp is an algebraic finite-dimensional complement of Fp in Fmn . For each k ∈ N, there exist un,k ∈ Fp and vn,k ∈ Lp such that

~~zn,k = un,k + vn,k.~~

~~Since n ∈ N is ﬁxed, simplify the notations by setting~~

~~zk := zn,k,uk := un,k,vk := vn,k,y:= yn.~~

If all vk = 0, the claim is trivial. Therefore we may assume that the sequence (vk)k contains a maximal linearly independent subset {e1,e2,...,er }. Deﬁne a norm on the linear span span{e1,e2,...,er } by the formula λj ej := |λj |. j≤r j≤r

Let tj be the ﬁrst zk for which vk = ej . Clearly, tj − ej ∈ Fp, and for each k ∈ N = = there exist unique scalars λk,j such that vk j≤r λk,j ej . Note that from uk zk − vk ∈ Fp it follows that wk := zk − λk,j tj = zk − vk + λk,j (ej − tj ) ∈ Fp. j≤r j≤r

~~→ → → Since vk 0, j≤r λk,j ej 0. Hence j≤r λk,j tj 0, and since | |≤ ∈ N j≤r λk,j 1 for almost all k (by the deﬁnition of the norm . ), we deduce that~~

~~wk ∈ 2ac{zk : k ∈ N} for almost all k ∈ N. Therefore~~

~~wk ∈ Fp ∩ 2ac{zk : k ∈ N} for almost all k ∈ N. Finally, lim wk = lim zk − λk,j tj = lim zk − 0 = y. k j≤r 326 15 Sequential Properties in the Class G~~

The claim is proved. In fact, we showed that {wn,k : k ∈ N}⊂H ∩ Fp. Since x = ∈ ⊂ − ∈ limn yn pSp Fp and Fp has the property C3 as a metrizable space, we note x − −− = − = × − H . This shows that H H ,soE M ϕ has the property C3 . The space ϕ does not have the property C3 (see Proposition 15.6), so the proof is completed.

~~For (LB)-spaces, we have the following theorem.~~

~~Theorem 15.5 An (LB)-space E with its deﬁning sequence (En)n of Banach − spaces En has the property C3 if and only if E is isomorphic to some Em, to ϕ, or to the product Em × ϕ.~~

− ∈ N Proof Assume E has the property C3 . For each n ,letSn be the unit ball in the = Banach space En.LetBn be the closure of Sn in E. Clearly, E n span(Bn), and each set Bn is bounded in E. Moreover, (Bn)n is a fundamental sequence of bounded sets in E (see the proof of the last part of Theorem 15.3). If each span(Bn) is infinite- dimensional in span(Bn+1), then (as in the proof of Case 1 in Proposition 15.8) − we deduce that E does not have the property C3 , a contradiction. Therefore there exists m ∈ N such that N := span(Bm) is countable-codimensional in E.LetP be an algebraic complement of N in E. Then E = N ⊕ P topologically. Indeed, since any countable-codimensional subspace of a barrelled space is barrelled (see [328, Theorem 4.3.6]), the space N is barrelled and P is a topological complement endowed with the strongest locally convex topology; see also [352]. Applying the closed graph theorem [328, Theorem 4.1.10], we deduce that N is isomorphic to the Banach space Em.IfP is isomorphic to ϕ, then E is isomorphic to EM ×ϕ.IfEm is finite-dimensional, E is isomorphic to ϕ. Now assume that P is finite-dimensional. Then there exists n>msuch that P ⊂ En. Consequently, E = En, and the closed graph theorem is applied again to show that this equality is topological. For the converse implication, it is enough to apply Proposition 15.9. Chapter 16 Tightness and Distinguished Fréchet Spaces

Abstract In this chapter, we apply the concept of tightness to study distinguished Fréchet spaces. We show that a Fréchet space is distinguished if and only if its strong dual has countable tightness. This approach to studying distinguished Fréchet spaces leads to a rich supply of (DF )-spaces whose weak∗ duals are quasi-Suslin but not K-analytic. The small cardinals b and d will be used to improve the analysis of Köthe’s echelon nondistinguished Fréchet space λ1(A).

~~16.1 A characterization of distinguished spaces~~

Let E be a metrizable lcs with a decreasing basis (Un)n of absolutely convex neighborhoods of zero. Apart from typical dual topologies on E such as the strong topology β(E,E) or the Mackey topology μ(E,E), there is a natural way to topologize the space E ◦ using the inductive limit topology generated by a deﬁning sequence (span(Un ))n. := ◦ Indeed, let En span(Un ) be endowed with the Minkowski functional norm topology. Then each En is a Banach space, the sequence (En)n is increasing and = E n En.By(E ,i(E ,E)) we denote the space E endowed with the inductive limit topology i(E ,E) generated by the deﬁning sequence (En)n above. It is known (from Grothendieck) that i(E,E) is the bornological topology associated with the strong topology β(E,E)(i.e., β(E,E)is bornological if and only if i(E,E)= β(E,E);see[213, Theorem 13.4.2]). Dieudonné and Schwartz called a Fréchet space E distinguished if the strong dual (E,β(E,E)) is barrelled. Grothendieck observed that E is distinguished if and only if (E,β(E,E)) is bornological. In fact, it is known that for a metrizable lcs E the dual (E,β(E,E))is quasibarrelled if and only if (E,β(E,E))is barrelled if and only if (E,β(E,E)) is bornological; see [288, Proposition 25.12], [246, 29.4.(3)], [328, 8.3.44]. We call a metrizable lcs E distinguished if (E,β(E,E)) is quasibarrelled. Therefore, E is distinguished if and only if (E,β(E,E)) is bornological (i.e., β(E,E) = i(E,E)). A classical result of Grothendieck states that every (DF )-space for which every bounded set is metrizable is quasibarrelled; see [246, 29.3.12 (b)]. Hence:

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 327 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_16, © Springer Science+Business Media, LLC 2011 328 16 Tightness and Distinguished Fréchet Spaces

(*) A metrizable lcs E whose strong dual (E,β(E,E)) has all bounded sets metrizable is distinguished. The first example of a nondistinguished Fréchet space was described by Gro- thendieck and Köthe; it was the Köthe echelon space λ1(A) for the Köthe matrix A = (an)n defined on the set N × N by the formula an(i, j) := j for i ≤ n and an(i, j) := 1 otherwise. For more examples of nondistinguished Fréchet spaces, we refer to [62]. The distinguished Köthe echelon spaces were intensively studied by many specialists; see, for example, [45], [51], [52], [54], [56], [57], [157], [159], [158]. Taskinen [395] provided a concrete Fréchet space C(R) ∩ L1(R) (endowed with the intersection topology). A simple argument to this effect was presented in [64]. The argument below, due to Bierstedt and Bonet [55, Theorem 1], is also valid for the space C∞(Ω) of infinitely differentiable functions on an open subset Ω ⊂ RN endowed with the compact-open topology for the functions and each of their derivatives.

Proposition 16.1 The intersection space E := C∞(Ω) ∩ L1(Ω) is a nondistinguished Fréchet space. := | | ∈ Proof Let p0(f ) Ω f dμ for f E. Choose an increasing sequence (Kn)n of compact sets covering Ω such that every compact subset of Ω is contained in some Kn and the interior of Kn+1 \ Kn is nonvoid for each n ∈ N. The topology of E is deﬁned by the increasing sequence (pn)n of seminorms

α pn(f ) := p0(f ) + max max |f (x)| |α|≤n x∈Kn for each f ∈ E. We prove that (E,β(E,E)) is not bornological. It is enough to show that i(E,E) = β(E,E). We show that for each bounded set B ⊂ E there exists u ∈ B◦ such that for each n ∈ N there exists fn ∈ E with pn(fn) ≤ 1 and u(fn) = 2. This is enough since then V := {v ∈ E :|v(f )|≤pn(f ), f ∈ E} n will be a neighborhood of zero in i(E,E)and not in β(E,E). So, ﬁx a bounded set B in E. By the boundedness, for each n ∈ N there exists Mn > 0 such that pn(f ) ≤ Mn for all f ∈ B. For each n ∈ N, choose a compact set In with a nonempty interior in Kn+1 \ Kn such that the Lebesque measure μ(In)< − 2−n−1M 1 . Set u(f ) := 2 fdμfor all f ∈ E. Since |u(f )|≤2p (f ),we n+1 n In 0 have u ∈ E. On the other hand, |u(f )|≤2 μ(In) max |f(x)|≤2 μ(In)Mn+1 ≤ 1. x∈K + n n 1 n 16.1 A characterization of distinguished spaces 329

◦ Hence u ∈ B . For each n ∈ N, choose a nonnegative test function fn on In such that f dμ = 1. Then f ∈ E for each n ∈ N. Each f vanishes on an open neigh- In n n n borhood of Kn. Then pn(fn) = p0(fn) = 1 and u(fn) = 2.

If E is a nondistinguished metrizable lcs, the topologies i(E,E) and β(E,E) are different. Even in that case, it may happen that (E,i(E,E)) = (E,β(E,E)). Indeed, there exists a nondistinguished Fréchet space E such that (E,i(E,E)) = (E,β(E,E)). The ﬁrst example of this type was provided by Komura;¯ see [45]or [421]. On the other hand, Grothendieck [188] showed that for the nondistinguished Köthe echelon space E := λ1(A) there exists a discontinuous linear functional on (E,β(E,E))that is bounded (i.e., transforms bounded sets to bounded sets, which means (E,i(E,E)) = (E,β(E,E))). Valdivia [421] proved that if E is a separable Fréchet space not containing a copy of the space 1, then (E,i(E,E)) = (E,β(E,E)). This surely motivates the question of whether every separable Fréchet space not containing a copy of 1 is necessarily distinguished. This problem has been answered in the negative by Diaz [115]. In order to prove Theorem 16.1 below, we need a couple of additional facts about the vector-valued Fréchet space 1(E), where E is a Fréchet space. The strong dual 1 := 1 1 1 (E)b (E) ,β( (E) , (E)) ˆ is isomorphic to the space (1⊗π E) . The latter space is isomorphic to the space ∞ b (Eb) of all bounded sequences in Eb endowed with the topology of uniform := convergence. The duality is deﬁned by the map u(x) i < x(i), u(i) >, where = ∈ 1 = ∈ ∞ x (x(i))i (E) and u u(i)i (Eb);see[330, Theorem 1.5.8]. The proof of (i) ⇔ (iii) in Theorem 16.1 is adopted from [55, Theorem 10].

Theorem 16.1 (Bierstedt–Bonet) For a Fréchet space E, the following conditions are equivalent: (i) E satisﬁes the density condition. (ii) Every bounded set in (E,β(E,E))is metrizable. (iii) The space 1(E) is distinguished.

Proof Note that the equivalence (i) ⇔ (ii) was already proved in Proposition 6.16. Let (Un)n be a decreasing basis of absolutely convex neighborhoods of zero in E. := ◦ Then the polars Bn Un form a fundamental sequence of bounded sets in the := strong dual Eb (E ,β(E ,E)). Consequently, the sets := { ∈ ∞ : ∈ ∈ N} Dn u (Eb) u(i) Bn,i ∞ form a fundamental sequence of bounded sets in the (DF )-space (Eb). ⇒ 1 ∞ (i) (iii): Since ( (E))b is isomorphic to (Eb), it is enough to show that ∞(E ) is bornological. Let D be an absolutely convex set in ∞(E ) that absorbs b ∞ b bounded sets in (Eb). Then there exists a sequence (λj )j of positive numbers 330 16 Tightness and Distinguished Fréchet Spaces such that n H := λj Dj ⊂ D. n j=1 ∞ Since (Eb) is a (DF )-space, to show that H is a neighborhood of zero it is enough to prove that H ∩ Dn is a neighborhood of zero in Dn for each n ∈ N [246, 29.3.(2)]. Then we will conclude that D is a neighborhood of zero. E satisfies the density condition (see Proposition 6.16). For the sequence (λj )j (chosen above) and each n ∈ N, by using the bipolar theorem, we find m>nand a bounded set B ⊂ E such that ◦ Bn ∩ B ⊂ λj Bj . 1≤j≤m Set := { ∈ ∞ : ∈ ◦ ∈ N} V u (Eb) ui B ,i . ∞ Then V is a neighborhood of zero in (Eb).Also Dn ∩ V ⊂ λj Bj ⊂ H. 1≤j≤m The last inclusion yields the conclusion. ⇒ ∞ (iii) (i): Assume E does not satisfy the density condition and (Eb) is bornological (i.e., 1(E) is distinguished). By the bipolar theorem, there exist a sequence (λj )j of positive numbers and n ∈ N such that for each m and each bounded ⊂ ∩ ◦ := set B E the set Bn B is not contained in Dm ac( 1≤j≤m λj Bj ). For each m ∈ N, define ∞ Am := {u ∈ (E ) : u(i) ∈ Dm,i∈ N}. b ∞ Note that A := Am is absolutely convex and absorbs bounded sets in (E ). m ∞ b Therefore A is a neighborhood of zero in the bornological space (Eb). Hence there exists a bounded set B ⊂ E such that := { ∈ ∞ : ∈ ◦ ∈ N}⊂ T u (Eb) u(i) B ,i A. ◦ On the other hand, for B and for each m ∈ N there exists u(m) ∈ (Bn ∩ B ) \ Dm. Also u = (u(m))m ∈ T ⊂ A. Hence there exists k ∈ N such that u ∈ Ak,sobythe definition of the set Ak we have that u(k) ∈ Dk, a contradiction.

By (*), any condition from Theorem 16.1 implies that E is distinguished. The converse fails in general, as the following observation shows. It is known that every separable (DF )-space is quasibarrelled; see [328, Proposition 8.3.13]. There exist reﬂexive Fréchet spaces whose strong dual is separable (such spaces are distinguished) and that do not satisfy the density condition [51]. It turns out that for the Köthe echelon spaces λ1 the density condition characterizes the distinguished property of λ1;see[51]. 16.1 A characterization of distinguished spaces 331

~~Theorem 16.2 The Köthe echelon space λ1 is distinguished if and only if it satisﬁes the density condition.~~

Proof Assume λ1 satisﬁes the density condition. Applying Theorem 16.1, we de- 1 1 ˆ duce that (λ1) is distinguished. Since (λ1) is isomorphic to the space 1⊗π λ1, the space λ1 is distinguished; see [51, Proposition 3]. Conversely, assume that λ1 is distinguished. In order to prove that λ1 satisﬁes 1 the density condition, it is enough to show that (λ1) is distinguished and apply ˆ Theorem 16.1. To see that 1⊗π λ1 is distinguished, we refer the reader again to the article [51, Proposition 3].

:= := −1 Let V (vn)n be the associated decreasing sequence on I (i.e., vn an for all n ∈ N, where A := (an)n is a strictly positive Köthe matrix on I ; in other words, an increasing sequence of strictly positive functions an on I ). Proposition 16.2 provides the condition (D) due to Bierstedt and Meise [56], which characterizes the density condition for λp,1≤ p<∞,orp = 0(see[54, Theorem 3]).

Proposition 16.2 An echelon space λp := λp(I, A),1≤ p<∞ or p = 0 satisﬁes the density condition if and only if A = (an)n satisﬁes the condition (D) (independent of p). In other words, there exists an increasing sequence (In)n of subsets of I such that ∈ N ∈ N −1 ≥ (i) for each m there is n(m) with infi∈Im an(m)(i)(ak(i)) > 0 for k n(m), while (ii) for each n ∈ N and each J ⊂ I with J ∩ (I \ Im) nonempty for all m ∈ N −1 there is k = k(n,J) > n with infi∈J ak(i)(an(i)) = 0.

This can be used to define a sufficient condition (denoted by (ND)) for the nondis- tinguishedness of λ1;see[55] for many discussions concerning this condition and its consequences. We provide another characterization for distinguished spaces. First we prove two additional propositions. The first one is motivated by the proof of Proposition 15.3.

Proposition 16.3 Let τ and ϑ be two locally convex topologies on a vector space E such that τ ≤ ϑ and each countable ϑ-equicontinuous set is equicontinuous in τ . If A is a ϑ-closed set and has τ -countable tightness, then τ|A = ϑ|A.

Proof Assume, by contradiction, that τ|A is different from ϑ|A. Then there exists a ϑ|A-closed (hence ϑ-closed) subset B of A that is not τ|A-closed. Let B denote the τ|A-closure of B. There exists x ∈ A such that x ∈ B \ B. Hence there exists a ϑ-continuous seminorm p on E such that p(x − y) ≥ 1 for each y ∈ B.Weshow that x/∈ {xn : n ∈ N} (the closure in τ restricted to A) for each sequence (xn)n in B, contradicting the τ -countable tightness of A.Let(xn)n be a sequence in B.ByF denote the linear hull of the set {x − xn : n ∈ N}. For each n ∈ N, choose a linear functional gn on F such that

gn(x − xn) = p(x − xn) ≥ 1, |gn(y)|≤p(y), 332 16 Tightness and Distinguished Fréchet Spaces for all y ∈ F . For each n ∈ N,lethn be a linear extension of gn to the whole E such that |hn(z)|≤p(z) for each z ∈ E. Deﬁne

−1 fn(z) := hn(z)(p(x − xn)) for each n ∈ N. Clearly, each fn is ϑ-continuous and fn(x −xn) = 1, |fn(z)|≤p(z), for each z ∈ E.If >0 and U ={z ∈ E : p(z) < }, the set U is a ϑ-neighborhood of zero in E such that |fn(z)|≤p(z) < for all z ∈ U and n ∈ N. Hence (fn)n is a ϑ-equicontinuous sequence and consequently it must be τ -equicontinuous (by the assumption on E). Therefore there exists an absolutely convex τ -neighborhood of −1 zero V such that |fn(z)| < 2 for all z ∈ V , n ∈ N and

~~(x + V)∩{xn : n ∈ N}=∅.~~

~~This provides a contradiction.~~

~~The next proposition extends Grothendieck’s result [246, 29.3.12(b)].~~

Proposition 16.4 Let E = (E, τ) be a (DF )-space. (i) E is quasibarrelled if and only if every bounded set in E has τ -countable tightness. (ii) If the Mackey space (E, μ(E, E)) is quasibarrelled, then (E, σ (E, E)) has countable tightness. If every bounded set in E has σ(E,E)-countable tightness, then (E, μ(E, E)) is quasibarrelled.

Proof Let (Sn)n be a fundamental sequence of absolutely convex, closed, bounded sets in E. (i): If E is quasibarrelled, by Theorem 12.3 the space E has countable tightness, and the conclusion follows. To prove the converse, assume that every bounded set in E has countable tightness. Since every linear functional on E that is continuous on each Sn is continuous on E, we apply (the proof of) Theorem 12.2 (i) ⇒ (ii) and Proposition 12.1 to deduce the following claim.

~~Claim 16.1 The weak∗ dual (E,σ(E,E))is realcompact.~~

Now we show that (E, μ(E, E )) is quasibarrelled. Indeed, for every se- := ∈ NN := o quence α (nk) ,setBα k nkSk . Since E is a (DF )-space, every sequence in any Bα is equicontinuous. Hence Bα is relatively countably compact in (E ,σ(E ,E)). Since (E ,σ(E ,E))is realcompact, every Bα is relatively compact and thus μ(E, E)-equicontinuous. As every β(E,E)-bounded set is contained in some Bα, each β(E ,E)-bounded set is μ(E, E )-equicontinuous. This proves that (E, μ(E, E)) is quasibarrelled.

~~Claim 16.2 We have the equality τ = μ(E, E). 16.1 A characterization of distinguished spaces 333~~

Indeed, since (E, τ) is a (DF )-space, the assumption of Proposition 16.3 for ϑ := μ(E, E) is satisfied. Then τ|A = μ(E, E )|A (16.1) for every bounded set A of E. Since (E, τ) is a (DF )-space, the topology τ is the finest locally convex topology on E satisfying (16.1); see [246, 29.3.2]. This implies the conclusion. (ii): If (E, μ(E, E)) is quasibarrelled, then (E, σ (E, E)) has countable tightness. Indeed, the sets Bα defined above compose a G-representation for (E, μ(E, E )). Now it is enough to apply Theorem 12.3. Note that the same argument as above shows the remaining implication.

~~For the next result, we refer to [158].~~

~~Theorem 16.3 (Ferrando–Kakol–López-Pellicer) ˛ A metrizable lcs E is distinguished if and only if every bounded set in the strong dual of E has countable tightness.~~

~~Proof The proof follows from Proposition 16.4 since the strong dual of a metrizable lcs is a (DF )-space.~~

On the other hand, there are a lot of nonquasibarrelled spaces with countable tightness whose every bounded set is metrizable; the weak topology of every inﬁnite-dimensional Fréchet–Montel space has this property. Proposition 16.4 for (DF )-spaces Cc(X) can be read as follows.

Proposition 16.5 The following conditions are equivalent for a (DF )-space Cc(X): (i) The compact-open topology τc of Cc(X) is equivalent to the Banach topology [ ]:={ ∈ : | |≤ } generated by the unit ball X, 1 f Cc(X) supx∈X f(x) 1 . (ii) Every bounded set of Cc(X) has countable tightness. (iii) Every bounded set of Cc(X) has countable tightness in the weak topology.

Proof Recall that X is pseudocompact (see Proposition 2.26 or Theorem 2.14 (vii)). It is known that [X, 1] generates on Cc(X) a ﬁner Banach topology ϑ, τc ≤ μ(Cc(X), Cc(X) ) ≤ ϑ, and the weak dual of (Cc(X), τc) is locally complete (see Theorem 2.14 (vi)). Clearly, (i) ⇒ (ii). If (ii) holds, Proposition 16.4 is applied to show that τc is barrelled (since a quasibarrelled space E is barrelled if and only if (E,σ(E,E)) is locally complete ([213, Theorem 11.2.5(b)]). Then the closed graph theorem applied to the identity map I : (Cc(X), τc) → (Cc(X), ϑ) yields ϑ = τc (see [328, Theorem 4.1.10]). This proves (i). The same argument applies to show (i) ⇔ (iii). Indeed, if (i) holds, the weak topology of Cc(X) has countable tightness by Theorem 12.2. Conversely, by Propo- sition 16.4, the Mackey topology μ(Cc(X), Cc(X) ) is barrelled. Again the closed graph theorem is applied to deduce that ϑ = μ(Cc(X), Cc(X) ). 334 16 Tightness and Distinguished Fréchet Spaces

~~16.2 G-bases and tightness~~

Recall that the character of an lcs E, denoted by χ (E), is the smallest inﬁnite cardinality for a basis of neighborhoods of zero. An lcs is metrizable if and only if its character is countable. Given an inﬁnite cardinal number m, denote by Gm the class of those locally convex spaces E for which χ(E)≤ m. Note that Gm is stable by taking subspaces, quotients by closed subspaces, completions and products of no more than m spaces; see [88]. We need the following result from [88].

Y Proposition 16.6 Let X and Y be topological spaces, and let ψ : X → 2 be an usco compact-valued map such that Y = {ψ(x) : x ∈ X}. If the weight ω(X) is inﬁnite, we have:(i) The Lindelöf number (Y n) ≤ ω(X) for every n ∈ N; and (ii) if Y is, moreover, metric, then dens (Y ) ≤ ω(X).

~~Proof To prove (i), observe that for every n ∈ N the multivalued map ψn : Xn → n 2Y deﬁned by~~

~~n ψ (x1,x2,...,xn) := ψ(x1) × ψ(x2) ×···×ψ(xn) is usco compact-valued, and n n n Y = {ψ (x1,x2,...,xn) : (x1,x2,...,xn) ∈ X }.~~

Since ω(X) is inﬁnite, we note ω(Xn) = ω(X). Hence we only need to prove our case for n = 1. Take (Gi)i∈I as any open cover of Y . For each x ∈ X, the compact set ψ(x) is covered by the family (Gi)i∈I . Therefore there exists a ﬁnite subset I(x) ⊂ ∈ of I such that ψ(x) i∈I(x)Gi. By the upper semicontinuity, for each x X ⊂ ∈ ⊂ there exists an open set Ox X such that x Ox and ψ(Ox) i∈I(x)Gi. The ⊂ family (Ox)x∈X is an open cover of X, and therefore there is a set F X such that | |≤ = F ω(X) and X x∈F Ox ; see, for example, [146, Theorem 1.1.14]. Then Y = ψ(X)= ψ(Ox) = Gi. x∈F x∈F i∈I(x)

Hence (Gi)i∈I has a subcover of at most w(X) elements. To get (ii), assume Y is a metric space, and for every n ∈ N choose Fn ⊂ Y a maximal set of points the distance between any two of which is at least n−1. Then ∈ ∩ Fn is closed, each x X has a neighborhood U such that ψ(U) Fn is ﬁnite, | |≤ = ∞ and therefore Fn ω(X). Then F n=1 Fn is dense in Y , and thus we obtain dens (Y ) ≤ ω(X), which ﬁnishes the proof.

Y Corollary 16.1 Let X and Y be topological spaces. Let ψ : X → 2 be an usco compact-valued map such that Y = {ψ(x) : x ∈ X}. If ω(X) is inﬁnite and if ⊂ n ≤ ∈ N Y0 Y is a closed subspace, (Y0 ) ω(X) for every n . 16.2 G-bases and tightness 335

~~n n n ≤ n Proof Since Y0 is closed in Y ,wehave(Y0 ) (Y ), and then we apply Propo- sition 16.6.~~

~~We are ready to prove the following result [88].~~

Theorem 16.4 (Cascales–Kakol–Saxon) ˛ Let {Es : s ∈ S} be a family of lcs in the : → ∈ = class Gm. Let fs Es E be linear maps for s S, and let E s∈S fs(Es) be the locally convex hull of {fs(Es) : s ∈ S}. Then t(E) ≤ m and t(E, σ (E, E )) ≤ m if |S|≤m.

Proof For every s ∈ S, fix a basis Bs of absolutely convex neighborhoods of zero in Es such that |Bs|≤m. Observe that t(E, τ) ≤ m. It is enough to show that, if A ⊂ E and 0 ∈ A (the closure in E), there exists B ⊂ A with |B|≤m such that 0 ∈ B. The family B := ac fs(Us) : Us ∈ Bs,s∈ S s∈S is a basis of zero in E, and the family B0 := ac fs(Us) : Us ∈ Bs,s∈ S, S finite subset of S s∈S has at most m elements. For A ⊂ E and 0 ∈ A,set := { : ∩ ∩ = ∅ ∈ } B xU0 xU0 is a chosen point in U0 A if U0 A ,U0 B0 . Clearly, B ⊂ A, |B|≤m and 0 ∈ B. Indeed, if U ∈ B,wehaveA ∩ U = ∅. Hence ∈ ⊂ ∩ = ∅ ∈ ∩ there exists U0 B0 such that U0 U and U0 A . Consequently, xU0 B U and 0 ∈ B. Now we prove that t(E, σ (E, E)) ≤ m. Since (E, σ (E, E)) is a subset of the space Cp(E ,σ(E ,E)), it suffices to show that (E ,σ(E ,E))n ≤ m (16.2)

for each n ∈ N. Indeed, then t(Cp(E ,σ(E , E))) ≤ m by Theorem 9.9, and this implies t(E, σ (E, E )) ≤ m. Since E is topologically isomorphic to the quotient = space E ( s∈S Es)/H, then (E ,σ(E ,E))is linearly homeomorphic to a closed subspace of s∈S(Es,σ(Es,Es)). To prove (16.2), it is enough to show that s∈S(Es,σ(Es,Es)) is an usco compact-valued image of a space of the weight at most m and then apply Corol- lary 16.1.Fors ∈ S, consider Bs as a discrete space. The map

(E ,σ (E ,Es )) ψs : Bs → 2 s s defined by ◦ ψs(U) := U for every U ∈ Bs 336 16 Tightness and Distinguished Fréchet Spaces = { : ∈ } is usco compact-valued, and Es ψs(U) U Bs .Themap (E ,σ (E ,Es )) ψ : Bs → 2 s s s s defined by ψ((Us)s) := ψs(Us) s ∈ for (Us)s s Bs is compact-valued and usco (see also [128, Proposition 3.6]) and satisfies Es = ψ((Us)s) : (Us)s ∈ Bs . s s ≤ Finally, we get ω( s Bs) m by applying [146, Theorem 2.3.13]. Hence t(E, σ (E, E )) ≤ m.

~~Theorem 16.4 yields the following classical result due to Kaplansky.~~

~~Corollary 16.2 If E is an lcs, then t(E) ≤ χ(E),t(E, σ (E, E)) ≤ χ(E).~~

An lcs E is said to have a G-basis [157] if the condition (iii) in Lemma 15.2 is N satisﬁed (i.e., there exists a family {Uα ∈ N } of neighborhoods of zero in E such N that Uα ⊂ Uβ whenever β ≤ α in N and such that each neighborhood of zero in E contains some Uα). Every metrizable lcs admits a G-base, with each Uα determined by the ﬁrst coordinate of α. Also, if E has a G-basis, every linear subspace of E has a G-basis. If F is a closed linear subspace of E, the quotient space E/F also has a G-basis. Note also that the completion of E has a G-basis.

Proposition 16.7 (i) If (En)n is a sequence of lcs each having a G-basis, the space = ∞ E n=1 En has a G-basis. (ii) If (En)n is a deﬁning sequence for an lcs E eachhavingaG-basis, the inductive limit space E has a G-basis. (iii) If (En)n is a family of lcs each having a G-basis, the locally convex direct = ∞ sum E n=1 En has a G-basis. { k : ∈ NN} ∈ N Proof (i): Let Vγ ) γ be a G-basis in Ek for each k . Set

α1 ∞ = k × Vα Vα Ek k=1 k=α1+1 N N for each α = (αn) ∈ N . Clearly, the family {Vα : α ∈ N } is decreasing. Fix m ∈ N. N Let Vk be a neighborhood of zero in Ek, and let βk = (βk,n) ∈ N be such that V k ⊂ V for 1 ≤ k ≤ m. Set βk k

~~αn = max{m, β1,n,...,βm,n} 16.2 G-bases and tightness 337~~

~~N for n ∈ N. Then α = (αn) ∈ N for α1 ≥ m and α ≥ βk for 1 ≤ k ≤ m. Hence m ∞ m ∞ V ⊂ V k × E ⊂ V × E . α βk k k k k=1 k=m+1 k=1 k=m+1~~

N It follows that {Vα : α ∈ N } is a G-basis in E. (ii): Let (Nk)k be a partition of N into inﬁnite subsets. Let ψk : N → N be a N strictly increasing map such that ψk(N) = Nk for k ∈ N. If α, β ∈ N for α ≤ β, N N N we have α ◦ ψk ≤ β ◦ ψk for all k ∈ N. Note that the map ϕ : N → (N ) deﬁned by ϕ(α) = (α ◦ ψk) is injective. Moreover, ϕ is a surjective map. Indeed, for β = N N (βk) ∈ (N ) ,weset = −1 α(n) βk(ψk (n)) ∈ ∈ N : N → N → = { k : ∈ for n Nk and k . Then α for n α(n), and ϕ(α) β. Let Vγ γ N N } be a G-basis of absolutely convex neighborhoods of zero in Ek for each k ∈ N. For α ∈ NN,set ∞ V = V k , α α◦ψk k=1 ∞ := ∞ n { : ∈ NN} where k=1 Vk n=1 k=1 Vk. Then Vα α is a G-basis of neighborhoods of zero in E. Indeed, let V be a neighborhood of zero in E. Then, for every ∈ N ∞ ⊂ k there exists a neighborhood of zero Vk in Ek such that k=1 Vk V. Let β ∈ NN be such that V k ⊂ V with k ∈ N. Then β = (β ) ∈ (NN)N,soβ = ϕ(α) k βk k k for some α ∈ NN. Hence ∞ ∞ V = V k ⊂ V ⊂ V. α βk k k=1 k=1

~~N Thus {Vα : α ∈ N } is a G-basis of neighborhoods of zero in E. Part (iii) follows from (ii).~~

Uncountable products of spaces with a G-basis need not admit a G-basis since every lcs with a G-basis belongs to the class G and uncountable products are not in the class G by Corollary 11.1. There are many spaces of this type. By Lemma 15.2, every quasibarrelled space in the class G has a G-basis. On the other hand, many spaces in G do not have a G-basis. Note that there exists a large class of quasibarrelled spaces (not in the class G) with countable tightness that do not admit a G- basis. Indeed, let X be an uncountable metrizable compact space. The space Cp(X) is quasibarrelled (see [213, Theorem 11.7.3]), and by Proposition 12.2 Cp(X) is not in the class G,soCp(X) does not have a G-basis. On the other hand, since X is n compact, X is Lindelöf for any n ∈ N. Hence Cp(X) has countable tightness by Proposition 9.9. Some (DF )-spaces admit a G-basis and some do not. The next two examples provide G-bases for some (DF )-spaces. 338 16 Tightness and Distinguished Fréchet Spaces

(i) Every strict (LB)-space is a (DF )-space that by Proposition 16.7 has a G- basis. (ii) The strong dual E of a metrizable lcs F is a (DF )-space having a G-basis. Indeed, let (Vn)n be a decreasing basis of absolutely convex neighborhoods of zero ∈ NN = ◦ { : ∈ NN} for F . For every α ,setUα ( k nkVk) . Then Uα α is a G-basis in E.

~~16.3 G-bases, bounding, dominating cardinals, and tightness~~

The first part of this section deals with two results due to Saxon and Sanchez-Ruiz [357](seealso[72]) showing that: (a) The bounding cardinal b is the smallest infinite dimensionality for a metrizable barrelled space. (b) Every metrizable lcs of dimension less than b is spanned by a bounded set, and for any nonnormable metrizable lcs E the minimal size for a fundamental family of bounded sets in E is the dominating cardinal d. ℵ It is worth noticing here that, according to Mazur’s result (see [328]), 2 0 is the smallest infinite dimensionality for a Fréchet space. The second part of this section deals with the bounding and dominating cardinals to study spaces in the class G with a G-basis. N ∗ Recall that given α, β ∈ N with α = (ak)k and β = (bk)k, we write α ≤ β ∗ to mean that ak ≤ bk for almost all k ∈ N. Thus α ≤ β implies α ≤ β, but not conversely. It is easy to see that every countable set in (NN, ≤∗) has an upper bound; this fails for (NN, ≤). The bounding cardinal b and the dominating cardinal d, respectively, are defined as the least cardinality for unbounded and cofinal subsets of the quasiordered space (NN, ≤∗);see[349]. A subset C of NN is called cofinal if for each α ∈ NN there exists β ∈ C such that α ≤∗ β. A subset of NN is called unbounded (dominating) if it is unbounded (cofinal) in (NN, ≤∗). ℵ It is clear that in any (ZFC)-consistent system one has ℵ1 ≤ b ≤ d ≤ 2 0 . The continuum hypothesis (CH) requires all four of these cardinals to coincide. Yet it is (ZFC)-consistent to assume that any of the three inequalities is strict. Note also that scales (i.e., well-ordered cofinal subsets of (NN, ≤∗)) exist if and only if b = d [359, Remark, p. 144]. Define the relation equivalence =∗ on (NN, ≤∗) so that α =∗ β if and only if ∗ ∗ ∗ ak = bk for almost all k ∈ N. Thus α = β if and only if α ≤ β and β ≤ α. Let aˆ denote the equivalence class represented by α, and note that each aˆ is countable. For any metrizable lcs E, define db(E) as the least cardinality for funda- ℵ mental systems of bounded sets in E. It is known that db(E) ≤ 2 0 (see [409]). It is ℵ also known that for any nonnormable metrizable lcs E one has ℵ1 ≤ db(E) ≤ 2 0 . We show the following result from [359].

~~Proposition 16.8 If E is a nonnormable metrizable lcs, then db(E) = d. 16.3 G-bases, bounding, dominating cardinals, and tightness 339~~

Proof Let (Un)n be a (decreasing) basis of absolutely convex neighborhoods of zero N N in E.LetD be a d-sized dominating subset of N . For each α = (an) ∈ N , there exists β = (bn) ∈ D such that an ≤ bn for almost all n ∈ N. Set mα := (man)n and

~~D := {mα : m ∈ N,α∈ D}.~~

Then |D|=d =ℵ0d. Set B := anUn; α = (an) ∈ D . n Note that B is a fundamental family of bounded sets in E. Indeed, if B is a bounded set in E, for each n ∈ N there exists an ∈ N such that B ⊂ anUn. There exists β = (bn) ∈ D such that an ≤ bn for all n ∈ N. Hence B ⊂ anUn ⊂ bnUn ∈ B. n n Hence B is a fundamental family of bounded sets in E and

~~db(E) ≤|B|≤|D|=d.~~

To prove the converse, let B be a family of bounded sets in E. The space E is assumed to be nonnormed, so there exists a sequence (fn)n in E such that each fn(Un) is unbounded. For any B ∈ B and n ∈ N,letgB (n) be the smallest natural number such that

~~sup |fn(x)| an.Theset C := {xn : n ∈ N} is bounded, and for B ∈ B there exists n ∈ N with an >gB (n) such that~~

~~sup |fn(x)|≥|fn(xn)| >an >gB (n) > sup |fn(x)|. x∈C x∈B This shows that C is not in B,soB is not a fundamental family of bounded sets if |B| < d. Hence we have d ≤ db(E).~~

~~This easily yields the following well-known fact; see [213].~~

~~Corollary 16.3 If E is a metrizable lcs and has a fundamental sequence of bounded sets, the space E is normable.~~

~~Corollary 16.4 The minimal size for a basis of neighborhoods of zero for the strong dual of a nonnormable metrizable lcs is d. 340 16 Tightness and Distinguished Fréchet Spaces~~

~~We show that the cardinal b is the smallest inﬁnite dimensionality for metrizable barrelled spaces; see Theorem 16.5. We need the following three technical lemmas from [359] and [357].~~

~~N Lemma 16.1 There exists a dense barrelled subspace ψb of the space R such that dim (ψb) = b.~~

Proof Recall that for the ordinals α and β one has α<βif and only if α ∈ β.By N ui we denote the ith coordinate functional deﬁned on R . First we show that for N {rβ : β ∈ b}⊂N there exists

N {gβ : β ∈ b}⊂N N such that, if {sβ : β ∈ b}⊂R satisﬁes |sβ (n)|≤rβ (n) for all β ∈ b and n ∈ N, and ∈ N = nk = N if for all k one has vk i=1 tkiui with tknk 0 and nk+1 >nk in , then there exists α0 ∈ b such that

{vk(x) + uk(y) : k ∈ N} is unbounded if x,y ∈ RN are of the form x = gβ + aαgα,y= sβ + aαsα, α∈σ α∈σ for some finite subset σ of β with β>α0. Without loss of generality, we may assume that each rβ is an increasing function. Fix {fβ : β ∈ b}, an unbounded subset N of N with each fβ increasing. Assume that for β ∈ b we have already defined {gα : α<β}. Since b is minimal, let f be an upper bound in NN for fγ rβ + pαrα + pαgα : γ ∈ β, σ ⊂ β is finite,pα ∈ N . α∈σ α∈σ Without loss of generality, one can assume that f is strictly increasing. For each n ∈ N,set

~~gβ (1) := f(1), gβ (n + 1) := f(n+ 1) + 2gβ (n)f (n + 1).~~

~~Note that the scalar sequence (vk(x) + uk(y))k is unbounded. Indeed, choose {sβ : N β ∈ b} and (vk)k as required. There exists h ∈ N such that~~

nk−1 | |− | | h(j) tknk tki >j i=1 for each k ∈ N and nk−1 α0 and x 16.3 G-bases, bounding, dominating cardinals, and tightness 341 and y are of the form given previously, choose pα ∈ N such that |aα|≤pα for each α ∈ σ . Clearly, there exists M>0 such that ≥ ≥ + + gβ (n) f(n) fα0 rβ pαrα pαgα (n) α∈σ α∈σ for each n ≥ M. This yields −1 −1 |x(n + 1)[x(n)] |= gβ (n + 1) + aαgα(n + 1) gβ (n) + aαgα(n) α∈σ α∈σ −1 ≥ gβ (n + 1) − pαgα(n + 1) gβ (n) + aαpα(n) α∈σ α∈σ

−1 −1 ≥[gβ (n + 1) − f(n+ 1)][2gβ (n)] = 2gβ (n)f (n + 1)[2gβ (n)] = f(n+ 1) for all n ≥ M. Note that |x(n)|≤2gβ (n). As f(n+ 1) ≥ n + 1 for all n ∈ N, the preceding inequalities show that there exists L>Msuch that |x(n)|≥max1≤i≤n |x(i)| for all ≤∗ ∗ n>L. Since fα0 f and fα0 h, there exists j>Lsuch that ≥ f(j) fα0 (j) > h(j) and nk−1 h(j) and | | | | fα0 (nk) tknk >h(j)tknk > 1. Now we are ready to get the ﬁnal conclusion. nk | + |≥| |−| |= −| |≥ vk(x) uk(y) vk(x) uk(y) tkix(i) y(k) i=1

~~ n −1 k | | + − | |−| |≥ tknk gβ aαgα (nk) tkix(i) y(k) α∈σ i=1~~

~~ nk−1 | | − −| − | | |− + ≥ tknk gβ pαgα (nk) x(nk 1) tki rβ pαrα (k) α∈σ i=1 α∈σ~~

nk−1 | | + − − − − | |− tknk f(nk) 2gβ (nk 1)f (nk) pαgα(nk) 2gβ (nk 1) tki α∈σ i=1 + ≥| | − + + rβ pαrα (nk) tknk f fα rβ pαrα α∈σ α∈σ 342 16 Tightness and Distinguished Fréchet Spaces nk−1 + − | |− | | ≥ pαgα (nk) 2gβ (nk 1) f(nk) tknk tki α∈σ i=1 nk−1 nk−1 | |− | | ≥ | |− | | 2 f(nk) tknk tki 2 h(j) tknk tki > 2j. i=1 i=1

~~This shows that the sequence (vk(x) + uk(y))k is unbounded as claimed. Now let~~

ψb := F + span{ei : i ∈ N}, where F is the linear span of {gβ : β ∈ b} and ei , i ∈ N, are the unit vectors. Clearly N ψb is dense in R . The dual of ψb is span{ui : i ∈ N}. We prove that ψb is barrelled (in fact, Baire-like, since ψb is metrizable, see Corollary 2.4). Indeed, for {sβ : β ∈ b}={0} and an infinite-dimensional set {vk : k ∈ N} in E , we apply the first part of the proof to get α0 ∈ b such that {vk(x) : k ∈ N} is unbounded for x ∈ span{gβ : β ∈ b} having a component beyond α0.This ∗ shows that every bounded set T in the weak dual of ψb is finite-dimensional, and hence T is equicontinuous. This proves that ψb is barrelled.

~~Lemma 16.2 Let B be a family of bounded sets in a metrizable lcs E such that |B| < b. Then there exist scalars tB such that A := {tB B : B ∈ B} is bounded in E.~~

Proof Fix a basis (Un)n of absolutely convex neighborhoods of zero in E.IfB ∈ B and n ∈ N, there exists tB (n) ∈ N such that B ⊂ tB (n)Un. Since |B| < b, there exists N α = (an) ∈ N such that an ≥ tB (n) for each B ∈ B and almost all n ∈ N. Hence, for each B ∈ B we can ﬁnd 0

Lemma 16.3 Let A be a set in an lcs E with |A| < b. Let (fn)n ⊂ E such that fn(x) → 0 for each x ∈ A. Then there exists an increasing sequence (si)i in N with 1 ≤ si ≤ i + 1 for each i ∈ N such that si →∞and (sifi(x))i is bounded for each x ∈ A.

~~Proof For x ∈ A,set~~

~~−1 ax(n) := min{k ∈ N :|fi(x)|≤(n + 1) for all i ≥ k}.~~

N N Clearly, αx = (ax(n)) ∈ N . By the deﬁnition of b,wehaveβ = (bn) ∈ N such ∗ that αx ≤ β for all x ∈ A. Without loss of generality, one can assume that b1 = 1 and (bn)n is strictly increasing. The deﬁnition of αx ensures that for each x ∈ A 16.3 G-bases, bounding, dominating cardinals, and tightness 343

~~−1 one has |fi(x)|≤(n + 1) for all i ≥ ax(n) and n ∈ N. Then (n + 1)|fi(x)|≤1if bn ≤ i~~

~~{(n + 1)fi(x) : bn ≤ i~~

~~Lemma 16.3 provides the following corollary.~~

~~Corollary 16.5 (Saxon–Sanchez-Ruiz) Let E be an inﬁnite-dimensional normed barrelled space. The dimension of E is at least b.~~

Proof Assume that E has dimension less than b. Then, using Theorem 15.4 for the completion F of E, we obtain a sequence (fn)n in E = F such that fn(x) → 0for all x ∈ E and fn=1 for all n ∈ N. Lemma 16.3 is applied to ﬁnd a sequence (sn)n in N such that (snfn)n is σ(E ,E)-bounded and snfn→∞, a contradiction, since E is barrelled, so any σ(E,E)-bounded set must be equicontinuous.

~~Proposition 16.9 (Saxon–Sanchez-Ruiz) Any metrizable lcs of dimension less than b is spanned by a bounded set.~~

~~Proof The conclusion follows from Lemma 16.2.~~

~~We note the following result stronger than Corollary 16.5. The proof follows from Lemma 16.2 and Corollary 16.5.~~

~~Theorem 16.5 (Saxon–Sanchez-Ruiz) The least inﬁnite dimensionality of a metrizable barrelled space E is b.~~

Proof If E is normed, the conclusion follows from Corollary 16.5. If the nonnormable inﬁnite-dimensional metrizable barrelled space E has dimension less than b, then E contains a bounded barrel (Lemma 16.2). Hence E is a normed space, a contradiction.

In particular, note that ψb is not spanned by a bounded set. Indeed, if ψb is spanned by a bounded set B, it must be normable since the closed, absolutely convex hull L of B is a barrel in ψb,soL is a neighborhood of zero (which yields that ψb is normed). The second part of this section deals with spaces in the class G and cardinals b and d. First observe that nonmetrizable lcs’s with G-bases have precisely limited characters. We prove the following proposition [157].

~~Proposition 16.10 The character χ(E) of a nonmetrizable lcs E having a G-basis satisﬁes the condition b ≤ χ(E)≤ d. 344 16 Tightness and Distinguished Fréchet Spaces~~

N Proof Let {Uα : α ∈ N } be a G-basis for E. By the definition of d, there exists a cofinal set D in (NN, ≤∗) with |D|=d. Then the set Dˆ := {ˆα : α ∈ D} satis- ˆ N ˆ fies |D|=ℵ0 · d = d and is cofinal in (N , ≤), so that {Uβ : β ∈ D} is a basis of neighborhoods of zero of cardinality d, where βˆ is the countable equivalence class defined by β; see the text before Proposition 16.8. This implies that χ(E)≤ d. Since the cardinals are well ordered, there exists a subset A of NN with |A|=χ(E) such that {Uα : α ∈ A} is a basis of neighborhoods of zero in E. Observe that |A|≥b. Suppose that |A| < b. Then, by the definition of b, there is some β ∈ NN such that α ≤∗ β for every α ∈ A. Hence, for every α ∈ A there exists γ ∈ βˆ such that α ≤ γ , βˆ being the countable equivalence class defined by β; see the text before Proposi- ˆ tion 16.8. It follows that {Uγ : γ ∈ β} is a countable basis of neighborhoods of zero, a contradiction.

~~This and Corollary 16.2 yield the following corollary.~~

~~Corollary 16.6 If an lcs space E has a G-basis, the tightness of E and (E, σ (E, E)) is at most d.~~

For the convenience of the reader, recall again the example of a nondistinguished Fréchet space attributed to Grothendieck and Köthe [246, 31.7]. This is the vector space E := λ1 of all numerical double sequences x = (xij ) such that for each n ∈ N we deﬁne = (n) ∞ pn (x) aij xij < , i,j (n) = ≤ (n) = where aij j for i n and all j, ai,j 1fori>nand all j. The seminorms pn for n ∈ N generate a locally convex topology under which E is a Fréchet space. The dual E is identiﬁed with the space of double sequences u = (uij ) such that | |≤ (n) ∈ N uij caij for all i, j and suitable c>0,n . We show (using the concept of tightness) that the Köthe echelon space is indeed nondistinguished; see [157].

Theorem 16.6 (Ferrando–Kakol–López-Pellicer–Saxon) ˛ The tightness of the strong dual (E ,β(E ,E)) of the Köthe echelon space E := λ1 equals d, the dominating cardinal. Moreover, the tightness of (E,σ(E,E)) is between b and d.

~~Proof By t(E) we denote the tightness of (E,β(E,E)). For each f : N → N, f := f ∈ ∈ N deﬁne the double sequence v (vij ) E so that, for all i, j , 0ifj ≤ f (i) , vf = ij 1ifj>f(i) .~~

f Thus, if i is ﬁxed, the single sequence (vij )j consists of zeros for the ﬁrst f(i) coordinates and ones thereafter. Set A ={vf : f ∈ NN}. We prove the theorem in three steps. 16.3 G-bases, bounding, dominating cardinals, and tightness 345

Claim 16.3 The origin zero belongs to the β(E,E)-closure of A. Indeed, let B be an arbitrary bounded set in E, choose g ∈ NN such that g(i) is an upper bound for i f pi(B) for each i ∈ N and set f(i):= 2 · g(i), thus determining v ∈ A.

Let x := (xij ) be an arbitrary member of B. Then j x,vf ≤ x vf = x ≤ x ij ij ij f (i) ij i,j i≥1 j>f(i) i≥1 j>f(i) 1 1 1 ≤ j xij ≤ pi (x) ≤ = 1. f (i) f (i) 2i i≥1 j≥1 i≥1 i≥1

~~This proves that v ∈ A B◦. Thus A meets every β(E,E)-neighborhood of zero. Claim 16.3 is proved.~~

Claim 16.4 t(E) ≥ d. Indeed, it is enough to show that zero is not in the closure of any subset of A having fewer than d elements. Let C := {vf : f ∈ F}, where F is a subset of NN with |F| < d. By the definition of d,thesetF is not cofinal in (NN, ≤∗). Hence there exists h ∈ NN such that h ≤∗ f does not hold for every ∈ F ∈ N r = r ∈ f . For each r , define x (xij )ij E by the formula 2if (i, j) = (r, h (r)) , xr = ij 0if(i, j) = (r, h (r)) .

r r Note that D := {x : r ∈ N} is bounded in E since, for a given n,wehavepn(x ) = 2 for all r>n, implying that pn(D) is a ﬁnite, and hence bounded, set. Let f be an arbitrary member of F. Because h ≤∗ f fails, there exists some r ∈ N with h(r) > f (r). Therefore, r f = r f = · f = x ,v xij vij 2 vr,h(r) 2. i,j

~~Finally, we note that vf ∈/ D◦ and D◦ is a β(E,E)-neighborhood of zero in E that misses C.~~

~~Claim 16.5 t(E) ≤ d. This follows from Corollary 16.6 and (ii) at the end of Sec- tion 16.2.~~

Now we prove the second part of the theorem. Step I. Since the strong dual E of any metrizable lcs E has a G-basis, by Corol- lary 16.6 we have that t(E,σ(E,E)) d. Step II. We prove that t(E,σ(E,E)) b. Define the set A as in the proof of the previous part. The β(E,E)-closure of A contains zero, and so does its closure in the coarser topology σ(E,E). It is enough to prove that zero is not in the σ(E,E)-closure of any subset C := {vf : f ∈ F} of A with |F| < b.Bythe 346 16 Tightness and Distinguished Fréchet Spaces definition of b,thesetF is bounded in (NN, ∗). Hence there exists g ∈ NN such that f ∗ g for every f ∈ F. Thus, for each f ∈ F there exists m(f ) ∈ N such that f(n) g(n) for all n>m(f). If we define h ∈ NN such that each h(n) = g(n) + 1, then, for every f ∈ F,

~~h (n) >f(n) for all n>m(f ) .~~

Just as before, we deﬁne xr in terms of this h and note that D := {xr : r ∈ N} is bounded in E, so that its polar D◦ is a neighborhood of zero in E. Thus D, viewed canonically as a subset of E, is equicontinuous on E. The Alaoglu– Bourbaki theorem provides z ∈ E such that xr ∈ z + V for inﬁnitely many r ∈ N whenever V is a σ(E,E)-neighborhood of zero. For an arbitrary f ∈ F,set V ={u ∈ E :|u(vf )| < 1} and choose r>m(f)such that xr − z ∈ V . Then 2 − z vf 2 − z vf = xr ,vf − z vf = xr − z vf < 1, which implies that |z(vf )| > 1. Hence {z}◦ is a σ(E,E)-neighborhood of zero that excludes each vf ∈ C.

This theorem combined with Theorem 12.3 shows that the Köthe echelon space E = λ1 is nondistinguished. Valdivia [421, (24), p. 66] showed that for a Fréchet space E the second dual (E,σ(E,E)) is a K-analytic space if and only if (E,μ(E,E)) is barrelled. Consequently, (E,σ(E,E)) is K-analytic for any distinguished Fréchet space E. As another argument, if a Fréchet space E is distinguished, then (E,β(E,E)) is a quasibarrelled (DF )-space. Then, by Theorem 12.3, the space (E,σ(E,E)) has countable tightness. Consequently, using Theorem 12.2, we deduce that (E,σ(E,E)) is K-analytic. On the other hand, it is easy to deduce from [246, 29.4.(3)] that for a Fréchet space E every locally bounded (i.e., bounded on bounded sets) linear functional on (E,β(E,E))is continuous if and only if (E,μ(E,E)) is bornological. Komura¯ [244] constructed a nondistinguished Fréchet space E such that every locally bounded linear functional on (E,β(E,E)) is continuous. For nondistinguished Köthe echelon spaces λ1, the situation is different. Proposition 16.4 is applied to provide another (simpler) proof of the following deep result of Bastin and Bonet [45, Theorem 2].

~~Corollary 16.7 If λ1 is a nondistinguished Köthe echelon space, there exists on (λ1,β(λ1,λ1)) a locally bounded discontinuous linear functional.~~

~~Proof Since, by Theorem 16.6, the space (λ ,σ(λ ,λ)) does not have countable 1 1 1 tightness, we apply Proposition 16.4 to deduce that the Mackey space μ(λ1,λ1) is not quasibarrelled.~~

Valdivia [421] invented a nondistinguished Fréchet space whose weak∗ bidual is quasi-Suslin and not K-analytic. We prove that Köthe’s original nondistinguished Fréchet space provides the same effect. 16.3 G-bases, bounding, dominating cardinals, and tightness 347

Example 16.4 below deals with certain spaces Cc(X);see[157]. In fact, we will work with spaces Cc(κ) := Cc([0,κ)), where κ is an infinite ordinal. Morris and Wulbert [305] studied the space Cc(ω1), where ω1 is the first uncountable ordinal. The cardinal of [0,ω1) is the first uncountable cardinal ℵ1. The set [0,κ)of all ordinals less than κ is endowed with its usual interval topology. For each ordinal α, the closed interval [0,α] is compact. However, for κ an infinite ordinal, [0,κ)is not compact and has a fundamental system of compact sets consisting of the sets [0,α] as the ordinal α ranges over a cofinal subset A of [0,κ). Clearly, the compact-open topology for Cc(κ) has a basis of neighborhoods of zero described by sets of the form

~~−1 Un,α := {f ∈ C(κ) :|f(γ)|≤n ,γ ∈[0,α]}, where n ∈ N and α ∈ A. Since~~

|{Un,α : n ∈ N,α∈ A}| = ℵ0 ·|A|=|A|, we obtain that the character χ(Cc(κ)) equals the cofinality cf(κ). The cofinality cf(κ) of an infinite ordinal κ is the smallest cardinality for cofinal subsets of [0,κ), where S ⊂[0,κ)is cofinal if for each ordinal α<κ(i.e., for each α ∈[0,κ)) there exists β ∈ S such that α ≤ β. The space Cc(κ) is a Fréchet space or a (DF )-space provided cf(κ) =ℵ0 or cf(κ) > ℵ0, respectively. Indeed, it is easy to see that Cc(κ) is sequentially complete, and hence it is a Fréchet space if cf(κ) =ℵ0. Warner [413] proved that Cc(X) is a (DF )-space if and only if every countable union of compact sets in X is relatively compact. A countable union of compact sets in [0,κ) is contained in a countable union of closed intervals, and their right endpoints have supremum β<κif cf(κ) is uncountable. Therefore the countable union of compact sets is contained in a compact interval [0,β], so it is relatively compact. We prove the following proposition.

~~Proposition 16.11 t(Cc(κ)) = t(σ(Cc(κ), Cc(κ) )) = χ(Cc(κ)) = cf(κ).~~

Proof The set C of characteristic functions of the open intervals (α, κ) is a subset of Cc(κ) whose closure contains zero. If B is any subset of C of size less than cf(κ), the collection of left endpoints has the supremum β<κ, so that all members of B are identically one on the open interval (β, κ). Choose γ ∈ (β, κ). The evaluation functional δγ is in the dual Cc(κ) and bounds B away from zero. Hence zero is not in the closure of B in σ(Cc(κ), Cc(κ) ). This shows that the tightness of both the original and weak topologies for Cc(κ) is at least cf(κ). Now we apply Corollary 16.2 to deduce that the tightness of the original and weak topologies for Cc(κ) is at most cf(κ).

We collect a couple of examples providing more spaces with (and without) G-bases. Consider the Banach space p(Λ), where p is ﬁxed with 1 ≤ p<∞ and 348 16 Tightness and Distinguished Fréchet Spaces

Λ is an uncountable indexing set. Let D be the closed unit ball in p(Λ). For each S ⊂ Λ, we deﬁne p ES := {u ∈ (Λ) : u(x) = 0,x∈ /S}, and for each countable T ⊂ Λ and each n ∈ N, also deﬁne

~~−1 [n, T ]:=(n D) + EΛ\T .~~

By E we denote the space p(Λ) endowed with the locally convex topology ξ having as a basis of neighborhoods of zero all sets of the form [n, T ]. Note that, for each countable T , the subspaces ET and EΛ\T are topologically complementary in E, that ET inherits the same Banach topology from E as it does from the Banach space p(Λ) and that the dual of E is the same as that of p(Λ). Observe that E := (E, ξ) is a sequentially complete nonquasibarrelled (DF )- space that does not have countable tightness and whose weak∗ dual is K-analytic. Indeed, note that since each sequence in E is contained in a Banach subspace of E, the sequential completeness is also clear. Since ξ is compatible with the Banach ∗ topology, the weak dual of E is K-analytic. Also, (nD)n forms a fundamental sequence of bounded sets. Then, to prove that E is a (DF )-space, we need to check that E is ℵ0-quasibarrelled. Indeed, let (Un)n be a sequence of absolutely convex closed neighborhoods of zero whose intersection U is bornivorous in E. For each n ∈ N that has 0 in its closure and is not in the closure of any countable subset of B) there exists a countable Tn ⊂ Λ and kn∈ N such that [kn,Tn]⊂Un. Therefore, ⊂ ⊂ := EΛ\Tn Un and EΛ\T U, where T n Tn. Since ET is a Banach space, it is ℵ0-barrelled, and U ∩ET is a relative neighborhood of zero. Since U intersects both summands ET and EΛ\T in neighborhoods of zero, it is a neighborhood of zero in E. This ensures that E is a (DF )-space, and it is clear that E is not Mackey since it is not the Banach space p(Λ). Therefore E cannot be quasibarrelled, nor can it have countable tightness by Proposition 16.4. (As alternative proof, the set B of all characteristic functions of the singleton subsets of Λ has 0 in its closure but not in the closure of any countable subset of B.) Note that E ∈ G for every choice of uncountable Λ, but we will prove that E admits a G-basis only when Λ is restricted under an axiomatic assumption milder than (CH).

~~Example 16.1 If ℵ1 = b =|Λ|, then E has a G-basis.~~

Proof By the definition of b, there is an injective map ϕ from [0, b) onto a set A unbounded in (NN, ≤∗), and by the assumption there is an injective map ψ from Λ onto [0, b). N For arbitrary σ = (a1,a2,...)∈ N ,letβ(σ) be the first member of [0, b) such ∗ that ϕ(β(σ)) (a2,a3,...), and define the corresponding neighborhood Uσ of zero by −1 Uσ := [a1,ψ ([0,β(σ)])]. −1 Note that β(σ) < b =ℵ1 implies the set [0,β(σ)] is countable. Thus ψ ([0,β(σ)]) is a countable subset of Λ, and hence Uσ is a neighborhood of zero in E. Clearly, 16.4 More about the Wulbert–Morris space Cc(ω1) 349

σ ≤ τ implies β(σ) ≤ β(τ) and Uτ ⊂ Uσ . To justify this inclusion, we need to N compare the ﬁrst coordinates. Finally, {Uσ : σ ∈ N } is a G-base because for a given n ∈ N and a countable T ⊂ Λ we set α = sup ψ(T) and note that α

~~Example 16.2 If ℵ1 < b, then E does not admit a G-basis.~~

Proof If E has a G-basis, then ES has a G-basis, where S is a subset of Λ of size ℵ1. Since the character of ES is ℵ1, we reach a contradiction with Proposition 16.10 and the assumption that ℵ1 < b.

~~Example 16.3 If |Λ| < b or |Λ| > d, then E does not admit a G-basis.~~

Proof Always ℵ1 ≤|Λ|, so the inequality |Λ| < b would imply ℵ1 < b, and the conclusion follows from the previous example. Now suppose that |Λ| > d and that N N ∗ {Uα : α ∈ N} is a G-basis for E.LetD be a cofinal subset of (N , ≤ ) of size d. ˆ N Then D := {ˆα : α ∈ D} is cofinal in (N , ≤) and still of size ℵ0 · d = d. The co- ˆ N ˆ finality of D in (N , ≤) ensures that U := {Uα : α ∈ D} is a base of neighborhoods of zero in E. We may choose a fixed t ∈ Λ \ {S(α) : α ∈ Dˆ } since this last union has size ℵ0 · d = d due to the countability of the sets S(α) used in the definition of Uα and d < |Λ|. Thus

~~χt ∈ EΛ\S(α) ⊂ Uα for each α ∈ Dˆ , contradicting the fact that U is a basis of neighborhoods of zero in the Hausdorff space E.~~

~~16.4 More about the Wulbert–Morris space Cc(ω1)~~

It is interesting to know when precisely the tightness of the space Cc(X) is countable; the same problem for Cp(X) has been discussed and solved in previous chapters. Lemma 16.4 will be used to show that the space Cc(ω1) does not have countable tightness. We shall say that an open cover Σ of X is compact-open if every compact subset of X is contained in some member of Σ. The following fact is due to McCoy. 350 16 Tightness and Distinguished Fréchet Spaces

~~Lemma 16.4 For any completely regular Hausdorff topological space X, the space Cc(X) has countable tightness if and only if every compact-open cover of X has a compact-open countable subcover.~~

Proof Assume that every compact-open cover of X has a countable compact-open subcover, and assume f ∈ A, where the closure is taken in Cc(X).LetK(X) be the family of all compact subsets of X. For every compact K ∈ K(X) and every n ∈ N, choose −1 fK,n ∈ A ∩{g ∈ Cc(X) :|f(x)− g(x)|

Then f ∈ B. Conversely, assume that Cc(X) has countable tightness, and let Σ be a compact-open cover of X. For every compact K in X, choose UK ∈ Σ such that K ⊂ UK . For every n ∈ N and every compact K ⊂ X, choose fK,n ∈ Cc(X) such −1 that n ≤ fK,n ≤ n and

~~−1 fK,n(K) ={n },fK,n(X \ UK ) ={n}.~~

~~Deﬁne~~

~~A := {fK,n : n ∈ N,A∈ K(X)}.~~

~~Since the closure of A contains the zero function 0, there are sequences (nj )j in N and (Kj )j in K(X) such that~~

~~∈ { : ∈ N} 0 fKj ,nj n, j . { : ∈ N} It is easy to see that UKj j is a compact-open subcover of Σ.~~

~~We apply Lemma 16.4 to get the following proposition.~~

~~Proposition 16.12 The Morris–Wulbert space Cc(ω1) does not have countable tightness.~~

Proof Note that Σ := {[0,α): α<ω1} is a compact-open cover of [0,ω1).IfK ⊂ [0,ω1) is compact, sup K<ω1. Assume that there exists a countable compact-open subcover of Σ. Then there exists an injective sequence (αn)n (i.e., αn = αm if n = m) of countable ordinals such that Γ := {[0,αn) : n ∈ N} is a compact-open cover := ∈ of Ω. Clearly, α supn αn <ω1. If γ (α, ω1), then (αn)n is contained in the 16.4 More about the Wulbert–Morris space Cc(ω1) 351 compact interval [0,γ]. By the assumption, there exists m ∈ N such that [0,γ]⊂ [0,αm), a contradiction since αm ∈[0,γ]\[0,αm).

~~We show that the existence of a G-basis in Cc(ω1) depends on the (ZFC)- consistent axiom system. We refer to results in [157].~~

~~Proposition 16.13 If ℵ1 < b, then Cc(ω1) does not admit a G-basis.~~

~~Proof Since the character of Cc(ω1) is ℵ1, Proposition 16.10 applies.~~

~~In the next proposition, as usual, b(d) denotes the ﬁrst ordinal of cardinality b(d).~~

~~Proposition 16.14 Both spaces Cc(b) and Cc(d) have a G-basis. If the ordinal κ has the coﬁnality ℵ0, b or d, then Cc(κ) has a G-basis.~~

N Proof For the space Cc(b), choose A ⊂ N with |A|=b such that A is unbounded in (NN, ≤∗).Letϕ be an injective map from [0, b) onto A. For arbitrary N σ = (a1,a2,...)∈ N , we deﬁne the corresponding neighborhood Uσ of zero by := [ ∗ Uσ Ua1,β , where β is the ﬁrst member of 0, b) such that ϕ(β) (a2,a3,...). Obviously,

~~σ ≤ τ ⇒ Uτ ⊂ Uσ . N Note that {Uσ : σ ∈ N } is a G-basis. Indeed, given n ∈ N and an ordinal α~~

~~∗ (a2,a3,...)≤ ϕ(β).~~

By the definition of d, one concludes that the result is a G-basis. Now we prove the second part: If cf(κ) =ℵ0, there exists a decreasing sequence (Vn)n that is a basis of neighborhoods of zero in Cc(κ).AG-basis is given by setting := U(a1,a2,... ) Va1 . If cf(κ) is b or d,letϕ be an injective map from a cofinal subset M of [0,κ) onto a subset A of NN such that |M|=b or d and A is unbounded or cofinal in (NN, ≤∗), respectively. Finally, follow the same procedure as before to construct a G-basis for Cc(κ).

~~Combining previous results, we obtain the following full characterization for the space Cc(ω1) to have a G-basis.~~

~~Proposition 16.15 The space Cc(ω1) has a G-basis if and only if ℵ1 = b.~~

Theorem 12.2, Theorem 12.3 and Proposition 16.12 imply that Cc(ω1) is a non- ∗ quasibarrelled (DF )-space and the weak dual of Cc(ω1) is not K-analytic. If we 352 16 Tightness and Distinguished Fréchet Spaces assume that ℵ1 = b, we note the following example also providing restrictions on possible extensions of Theorem 6.7.

~~Example 16.4 Set E := Cc(ω1). Then F := (E ,σ(E ,E)) is not K-analytic. If ℵ1 = b, the space F has a resolution of metrizable and compact, absolutely convex sets.~~

Proof Since every compact set in X := [0,ω1) is metrizable, the polar of every neighborhood of zero in E is σ(E,E)-metrizable by Proposition 6.8 and Lemma 6.5. It is also known that E is a locally complete (DF )-space and X is pseudocompact and noncompact, and also the family {f ∈ E : f(X)≤ 1} generates on E a Banach topology ϑ such that μ(E, E) ≤ ϑ;see[231] for more details. We show again (with a somewhat different argument) that (E,σ(E,E))is not a K-analytic space. Assume that F is K-analytic. Then σ(E,E) has countable tightness by Theorem 12.2, and (E, μ(E, E)) is quasibarrelled by Proposition 16.4. Hence the space (E, μ(E, E)) is barrelled (since E is a locally complete space). By the closed graph theorem applied to the identity map I : (E, μ(E, E)) → (E, ϑ), we have the equality μ(E, E) = ϑ.AsX is noncompact, the topological duals of E and (E, ϑ) are different; see [157, Lemma 2]. Hence (E,σ(E,E))indeed is not a K-analytic space. Now assume that ℵ1 = b. Then, by Proposition 16.15, we deduce that E has a N basis of absolutely convex neighborhoods of zero {Uα : α ∈ N } such that Uα ⊂ Uβ if β ≤ α. Clearly, the polars of the sets Uα compose a resolution consisting of σ(E,E)-metrizable compact absolutely convex sets.

From Fremlin’s theorem [346, Theorem 5.5.3], it follows that under (CH) there exists a nonanalytic K-analytic space E such that each compact set in E is metriz- ℵ able. It is known that in any (ZFC)-consistent system one has ℵ1 ≤ b ≤ 2 0 .Ifweas- sume Martin’s axiom and the negation of (CH), then any K-analytic space in which ℵ every compact set is metrizable is analytic [346, Theorem 5.5.3] and ℵ1 < b = 2 0 ; see, for example, [140]. We complete this section with some extra facts concerning the Morris–Wulbert space Cc(ω1). Recall, as Morris and Wulbert noted in [305], that Cc(ω1) is an ℵ0-space (i.e., it has a pseudobase) that is not barrelled and, in fact, is not even a Mackey space. Moreover, since [0,ω1) is pseudocompact and not compact, it follows that the dual of the Banach space Cu(ω1) with the sup-norm topology is strictly stronger than that of Cc(ω1) [157, Lemma 2], and thus the Mackey topology μ := μ(Cc(ω1,Cc(ω1) ) also is not barrelled, as the closed graph theorem shows. Let us compare directly the duals of Cc(ω1) and Cu(ω1). First, for each α ∈[0,ω1), set (α, ω1) := [0,ω1) \[0,α], and let

~~F ={f ∈ Cc(ω1) : f ((α, ω1)) ={0}, for some α ∈[0,ω1)}.~~

~~Let h be a function whose value is identically 1 on [0,ω1). 16.4 More about the Wulbert–Morris space Cc(ω1) 353~~

Claim 16.6 F is algebraically complemented in Cc(ω1) by the span of h. Indeed, if g is a real-valued function that is nonconstant on each set (α, ω1), there are uncountably many β ∈[0,ω1) such that |f(β)− f(β+ 1)| > 0, and thus there exists some n ∈ N such that |f(β)− f(β+ 1)| >n−1 for uncountably many β. Therefore we may choose β1 <β2 <... such that

−1 |f(βk) − f(βk + 1)| >n for all k ∈ N. Note also that the sequence (βk)k converges in [0,ω1), although (f (βk))k and (f (βk + 1))k do not converge to the same point in R. Therefore g is not continuous. We conclude that each f ∈ Cc(ω1) is eventually constant, so that Cc(ω1) is the algebraic direct sum of F and the span of h.

~~In the sup-norm, h is clearly unit distance from the subspace F and is in the compact-open closure of F .~~

Claim 16.7 F is a closed hyperplane in the space Cu(ω1), and it is a dense hyperplane in Cc(ω1). Therefore members of Cc(ω1) are uniquely determined by their restrictions to F . Let Fc and Fu denote F with its relative compact-open and uniform sup-norm topologies, respectively. Then Fu is a Banach space that dominates Fc.Given ∈ ∈[ = ∈ λ Fu, there exists a point α 0,ω1) such that λ(f ) 0 whenever f F with f([0,α]) ={0}. Indeed, otherwise, for some n ∈ N, there would be uncountably many α ∈[0,ω1) and f ∈ F with − f =1,λ(f)>n1,f([0,α]) ={0}.

But then we can ﬁnd f1,...,fk ∈ F having disjoint supports and unit norm, with −1 each λ(fj )>n .Takek>nλ. By virtue of disjoint supports, f1 +···+fk= 1, yielding the contradiction that

~~−1 kn <λ(f1 +···+fk) ≤f1 +···+fk=λ.~~

Having now established the existence of a point α ∈ ω1 such that λ(f ) = 0 whenever [ ] ={ } ∈ = ∈ f( 0,α ) 0 , we see that λ Fc and Fu Fc.Letϕ Cu(ω1) that vanishes = = on F and ϕ(h) 1. For any member λ of Fu Fc, there is a unique continuous linear extension λ0 to Cc(ω1), whereas each linear extension remains continuous on Cu(ω1) and may be realized in the form λ0 + cϕ, where c is an arbitrary scalar. We proved the following.

~~Claim 16.8 Cc(ω1) is a one-codimensional subspace of Cu(ω1) .~~

~~Finally, we prove the following statement.~~

Claim 16.9 Cc(ω1) is a closed hyperplane of the dual Banach space Cu(ω1) .In- deed, for any γ ∈ Cc(ω1) , there is some point α ∈[0,ω1) such that the characteristic function χα of [0,α] satisﬁes γ(χα) = γ(h). Otherwise, there would be an n ∈ N 354 16 Tightness and Distinguished Fréchet Spaces

−1 and an uncountable set Λ of points in ω1 such that |γ(h)− γ(χα)| >n for each α ∈ Λ. Then h is in the Cc(ω1)-closure of the set {χα : α ∈ Λ},butγ(h) is not in the closure of {γ(χα) : α ∈ Λ}, contradicting continuity of γ . With α ﬁxed such that γ(χα) = γ(h),wehave

~~γ + ϕ≥γ + ϕh − χα≥|(γ + ϕ)(h − χα)|=ϕ(h) = 1,~~

~~which shows that ϕ is unit distance from the hyperplane Cc(ω1) in the Banach space Cu(ω1)β , so that the hyperplane is closed.~~

Another approach might be the following: Since Cc(ω1) is a (DF )-space (see the text before Proposition 16.11), it is a (df )-space, and then its strong dual Cc(ω1)β is a Banach space by Theorem 2.14. It is known that the spaces Cc(ω1) and Cu(ω1) have the same bounded sets; see, for example, [363, Theorem 11]. Then the topology on Cc(ω1)β is that induced by Cu(ω1)β . Chapter 17 Banach Spaces with Many Projections

Abstract In this chapter, we discuss Banach spaces that have a rich family of norm- one projections onto separable subspaces. One of the tools, coming from logic, is the concept of an elementary submodel.

~~17.1 Preliminaries, model-theoretic tools~~

Recall that a projection in a Banach space E is a bounded linear operator P : E → E such that P 2 = P . A subspace Y of a Banach space E is complemented if there exists a projection P : E → E such that Y = P [E]. Equivalently, there exists a closed subspace Z of E such that Y ∩ Z = 0 and Y + Z = E.Asusual,wewrite E = Y ⊕ Z. One of the important tools for investigating nonseparable Banach spaces and, in particular, for constructing projections is the method of elementary substructures. Roughly speaking, given a Banach space X, we shall use a countable structure M such that XM := X ∩ M is a separable subspace of X that shares some properties of X or the embedding XM ⊂ X has a certain useful property. Elementary substructures are used successfully in set theory and topology, replacing various closing-off arguments. We shall now explain this concept. Let N be a fixed set. The pair (N, ∈), where ∈ is restricted to N × N,isa structure of the language of set theory. Given a formula ϕ(x1,...,xn) with all free variables shown and given a1,...,an ∈ N, one defines the relation “(N, ∈) satisfies ϕ(a1,...,an)” (briefly “(N, ∈) |= ϕ(a1,...,an)”orjust“N |= ϕ(a1,...,an)”) in the usual way, by induction on the length of the formula. Namely, N |= a1 ∈ a2 if and only if a1 ∈ a2 and N |= a1 = a2 if and only if a1 = a2. It is clear how “satisfaction” is defined for conjunction, disjunction and negation. Finally, if ϕ is of the form (∃ y)ψ(x1,...,xn,y), then N |= ϕ(a1,...,an) if and only if there exists b ∈ N such that N |= ψ(a1,...,an,b). For example, if s ={a,b,c} and s,a,b ∈ N while c/∈ N, then N satisfies “s has at most two elements” because, for every x ∈ N, if x ∈ s, then either x = a or x = b. Instead of the definition above, some authors use relativization; see, for example, Kunen’s book [258]. Given a formula ϕ,therelativization of ϕ to N is a formula ϕN that is built from ϕ by replacing each quantifier of the form “∀ x”by“∀ x ∈ N” and each quantifier of the form “∃ x”by“∃ x ∈ N.” In this way, N |= ϕ(a1,...,an) N if and only if ϕ (a1,...,an) holds (of course, a1,...,an must be elements of N).

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 355 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_17, © Springer Science+Business Media, LLC 2011 356 17 Banach Spaces with Many Projections = Given a set x, we define the transitive closure of x to be tc(x) n tcn(x), where tcn(x) = x ∪ x and tcn+1(x) = tc1(tcn(x)). In other words, y ∈ tc(x) if and only if there are x0 ∈ x1 ∈···∈xk such that y ∈ x0 and xk ∈ x. Thanks to the axiom of regularity, these two definitions of transitive closure are equivalent and every set of the form tc(x) is transitive (i.e., y ∈ tc(x) implies y ⊆ tc(x)). Given a cardinal θ, we denote by H(θ) the class of all sets whose transitive closure has cardinality <θ. It is well known that H(θ) is a set, not a proper class. It is clear that H(θ) is transitive. We shall consider elementary substructures of (H (θ), ∈). It is well known that, for a regular uncountable cardinal θ, the structure (H (θ), ∈) satisfies all the axioms of set theory, except possibly the power set axiom; see [258,IV.3]. Recall that a substructure M of (H (θ), ∈) is called elementary if for every formula ϕ(x1,...,xn) with all free variables shown, for every a1,...,an ∈ M,wehave that

~~M |= ϕ(a1,...,an) ⇐⇒ H(θ)|= ϕ(a1,...,an). The fact that M is an elementary submodel of (H (θ), ∈) is denoted by~~

~~M (H (θ), ∈).~~

The reason for using elementary submodels of H(θ) is that these structures satisfy most of the axioms of set theory: If θ>ℵ0 is regular, then H(θ)satisfies all the axioms except possibly the power set because it may happen that 2λ >θ for some λ<θ. Moreover, in practice it is usually easy to point out a cardinal θ that H(θ) satisfies given finitely many formulas with parameters needed for applications. Another useful feature of H(θ) with θ regular is the fact that for every formula ϕ(x1,...,xn) in which all quantifiers are bounded (i.e., of the form “∀ x ∈ y”or “∃ x ∈ y”) and for every a1,...,an ∈ H(θ), ϕ(a1,...,an) holds if and only if

~~H(θ)|= ϕ(a1,...,an).~~

For more information, see [258, IV.3]. Since in most cases we indeed use formulas with bounded quantifiers, one can simply “check” their validity by looking at a sufficiently large H(θ). One can also use the reflection principle, which says that given a formula of set theory ϕ(x1,...,xn) and given sets a1,...,an such that ϕ(a1,...,an) holds, there exists θ such that the structure (H (θ), ∈) satisfies ϕ(a1,...,an). In some cases, θ may not be regular, although it may be arbitrarily big and have arbitrarily big cofinality. More precisely, the class of cardinals θ with the property above is closed and unbounded. Thus, when considering finitely many formulas and parameters, we can “check” their validity by restricting attention to H(θ), where θ is a “big enough” cardinal, meaning that the cofinality of θ is greater than a prescribed cardinal and all relevant formulas are satisfied in (H (θ), ∈). Summarizing, assume we would like to use in our arguments formulas ϕ1,...,ϕn and parameters from a finite set S. We then find a cardinal θ such that S ⊆ H(θ)and, 17.1 Preliminaries, model-theoretic tools 357 by the reflection principle, all valid formulas ϕ1,...,ϕn with suitable parameters are satisfied in H(θ). Finally, we shall use elementary substructures of H(θ) that contain S. If the formulas ϕ1,...,ϕn have only bounded quantifiers, then we do not need to use the reflection principle since the formulas will be satisfied in every H(θ) with big enough θ (i.e., every regular θ greater than some fixed cardinal θ0). A particular case of the Löwenheim–Skolem theorem (for the language of set theory) says that for every infinite set S ⊆ H(θ) there exists M (H (θ), ∈) such that |M|=|S|. This theorem can be viewed as the “ultimate” closing-off argument, and its typical proof indeed proceeds by “closing-off” the given set S by adding elements that witness “satisfaction” of all suitable formulas of the form (∃ x) ψ.Im- portant for applications is the fact that, thanks to the Löwenheim–Skolem theorem, we may consider countable elementary substructures of an arbitrarily large H(θ).

~~Proposition 17.1 Let θ be an uncountable regular cardinal and M (H (θ), ∈).~~

(a) Assume u ∈ H(θ), a1,...,an ∈ M and ϕ(y,x1,...,xn) is a formula such that u is the unique element of H(θ) for which we have H(θ)|= ϕ(u,a1,...,an). Then u ∈ M. (b) Let s ⊆ M be a ﬁnite set. Then s ∈ M. (c) If S ∈ M is a countable set, then S ⊆ M.

Proof (a) By elementarity there exists v ∈ M such that M |= ϕ(v,a1,...,an).Using elementarity again, we see that H(θ)|= ϕ(v,a1,...,an). Thus u = v. (b) Let s ={a1,...,an}. Then s is the unique set satisfying ϕ(s,a1,...,an), where ϕ(x,y1,...,yn) is

~~(∀ t)t ∈ x ⇐⇒ t = y1 ∨ t = y2 ∨···∨t = yn.~~

Applying (a), we see that s ∈ M. (c) By induction and by (a), we see that all natural numbers are in M.Alsoby (a), the set of natural numbers ω is an element of M, being uniquely defined as the minimal infinite ordinal. Notice that H(θ)satisfies “there exists a surjection from ω onto S.” By elementarity, there exists f ∈ M such that M satisfies “f is a surjection from ω onto S.” Again using (a), we see that f(n)∈ M for each n ∈ ω. Finally, it suffices to observe that f is indeed a surjection (i.e., for every x ∈ S there is n such that x = f(n)). This follows from elementarity because, assuming f [ω]=S, the formula “(∃ x ∈ S)(∀ n ∈ ω) x = f(n)” would be satisfied in M, contradicting that f is a surjection.

Now fix a Banach space X and choose a “big enough” regular cardinal θ so that X ∈ H(θ). Take an elementary substructure M of (H (θ), ∈) such that X ∈ M.Note that M may be countable by the Löwenheim–Skolem theorem. What can we say about the set X ∩ M? By elementarity, it is closed under addition. By Proposition 17.1(a), the field of rationals is contained in M, and therefore X ∩ M is a Q-linear subspace of X. Consequently, the norm closure of X ∩ M is 358 17 Banach Spaces with Many Projections a Banach subspace of X. In particular, the weak closure of X ∩ M equals the norm closure of X ∩ M. We shall write XM instead of (X ∩ M) and shall call XM the subspace induced by M. In the case of some typical Banach spaces, we can describe the subspace XM . For instance, let X = p(Γ ), where 1 ≤ p<∞ and Γ is an uncountable set. Then XM can be identified with p(Γ ∩ M). Indeed, identify x ∈ p(Γ ∩ M) with its extension x ∈ p(Γ ) defined by x (α) = 0forα ∈ Γ \ M.Letx ∈ X ∩ M. Then

~~supp(x) ={α ∈ Γ : x(α) = 0} is a countable set and hence, by elementarity, it belongs to M. By Proposi- tion 17.1(c), we have supp(x) ⊆ M. Thus x ∈ p(Γ ∩ M). On the other hand, if~~

x ∈ p(Γ ∩ M), then arbitrarily close to x we can ﬁnd y ∈ p(Γ ∩ M) such that s = supp(y) is ﬁnite. Moreover, we may assume that y(α) ∈ Q for α ∈ s. By Proposition 17.1(b), y s ∈ M and consequently also y ∈ M. Hence

~~x ∈ (X ∩ M)= XM .~~

~~Given a compact space K ∈ H(θ) and M (H (θ), ∈), deﬁne the following equivalence relation ∼M on K:~~

~~x ∼M y ⇐⇒ (∀ f ∈ C(K) ∩ M)f(x)= f(y).~~

M We shall write K/M instead of K/∼M , and we shall denote by q (or, more pre- M cisely, qK ) the canonical quotient map. It is not hard to check that K/M is a compact Hausdorff space of weight not exceeding the cardinality of M. This construction has been used by Bandlow [43], [44] for characterizing Corson compact spaces in terms of elementary substructures.

~~Lemma 17.1 Let K be a compact space, let θ be a big enough regular cardinal and let M (H (θ), ∈) be such that K ∈ M. Then~~

~~C(K) ∩ M ={ϕ ◦ qM : ϕ ∈ C(K/M)}, where the closure is the norm closure in the formula above.~~

Proof Let Y denote the set on the right-hand side. Then Y is a closed linear subspace of C(K).Givenψ ∈ C(K)∩ M, by the deﬁnition of ∼M , there exists a (necessarily continuous) function ψ such that ψ = ψ ◦ qM . Thus C(K) ∩ M ⊆ Y .Let

~~R ={ϕ ∈ C(K/M): ϕ ◦ qM ∈ M}.~~

Then R is a subring of C(K/M) that separates points and contains all rational con- stants. By the Stone–Weierstrass theorem, R is dense in C(K/M), which implies that C(K) ∩ M is dense in Y . 17.1 Preliminaries, model-theoretic tools 359

~~Observe that, under the assumptions of the lemma above, the norm closure of C(K ∩ M) is pointwise closed. Indeed, if~~

~~f ∈ C(K) \ (K ∩ M), then there are x,y ∈ K such that x ∼M y, while f(x)= f(y). Consequently,~~

V ={g : g(x) = g(y)} is a neighborhood of f in the pointwise convergence topology, disjoint from (K ∩ M). Given a map of compact spaces f : K → L and given M H(θ) with f ∈ M, observe that the relation ∼M is preserved by M,

~~x ∼M y ⇒ f(x)∼M f(y).~~

Indeed, if f(x)∼M f(y)and ϕ ∈ C(L) ∩ M witnesses it, then ϕ ◦ f ∈ C(K) ∩ M separates x and y. As a consequence of this observation, for every map f : K → L in M there exists a (necessarily continuous) map f M : K/M → L/M for which the diagram

~~f K L~~

~~M M qK qL~~

~~K/M L/M f M~~

M M ◦ commutes. The continuity of f follows from the continuity of qL f and from M ◦ M = the fact that qK is a quotient map. It is an easy exercise to show that (f g) M M M M f ◦g whenever f,g ∈ M are compatible. Clearly, (id K ) = id K/M,sof → f is a functor from the category of compact spaces and continuous maps in M into the category of compact spaces of weight ≤|M|. Below we show that this functor preserves ﬁnite products.

Lemma 17.2 Let K,L be compact spaces, let θ be a big enough regular cardinal, and let M (H (θ), ∈) be such that K,L ∈ M. Then (K × L)/M is naturally homeomorphic to (K/M) × (L/M). The homeomorphism is given by the formula [ ] → [ ] [ ] (x, y) ∼M ( x ∼M , y ∼M ). (17.1)

M [ ] = Proof Let prK ,prL denote the respective projections. Then prK ( (x, y) ∼M ) [ ] M [ ] =[ ] x ∼M and prL ( (x, y) ∼M ) y ∼M . This shows that (17.1) well deﬁnes a continuous map that is clearly a surjection. It remains to show that it is one-to-one. Fix x ∼M x and y ∼M y . We must show that (x, y) ∼M (x ,y ). Suppose otherwise and choose ϕ ∈ M ∩ C(K × L) such that ϕ(x,y) < ϕ(x,y). Find rational numbers s,t such that ϕ(x,y)

ϕ−1[(←,s]] and B = ϕ−1[[t,→)] are disjoint closed subsets of K × L such that (x, y) ∈ A and (x,y) ∈ B. Furthermore, A,B ∈ M. Using compactness and elementarity, we may ﬁnd ﬁnite families U , V ∈ M if there are basic open subsets of K × L such that U covers A, V covers B and U ∩ V =∅.

Find U ∈ U and V ∈ V such that (x, y) ∈ U and (x ,y ) ∈ V . Then U = U0 × U1, V = V0 × V1 and U0,U1,V0,V1 ∈ M.Findi<2 such that Ui ∩ Vi =∅.By symmetry, we may assume that i = 0. Let A0 = U0 and B0 = V0. Then A0,B0 ∈ M are closed, disjoint and x ∈ A0, x ∈ B0. Using Urysohn’s lemma and elementarity, there is g ∈ M ∩ C(K) such that g A0 = 0 and g B0 = 1. Finally, g(x) = 0 and g(x ) = 1, which shows that x ∼M x .

~~The next statement shows that typical algebraic structures on a compact space are preserved under canonical quotients.~~

Corollary 17.1 Let K be a compact space, n ∈ ω, and let a : Kn → K be a continuous map. Furthermore, let M (H (θ), ∈), where θ is big enough and a ∈ M. Then M induces a continuous map aM : (K/M)n → K/M such that the canonical M M quotient qK becomes a homomorphism of the structures (K, a) and (K/M, a ).

~~Proof By Lemma 17.2, (K/M)n is homeomorphic to Kn/M and therefore aM is M ◦ = M ◦ M deﬁned as the unique map satisfying qK a a qKn . This equation actually shows that aM is a homomorphism.~~

~~Corollary 17.2 Let (K, <) be a compact line, let M H(θ) be such that θ is big enough, and (K, <) ∈ M. Then K/M is a compact line, and the canonical quotient qM : K → K/M is increasing.~~

Proof One can look at K as a meet semilattice (K, ∧), where x ∧ y = min{x,y}. By Corollary 17.1, K/M has a binary operator ∧M and qM is a homomorphism. Since it is surjective, it is clear that ∧M is a meet semilattice that deﬁnes a linear order by p ≤ q ⇐⇒ p = p ∧M q.

~~Lemma 17.3 Let K be a totally disconnected compact space, let θ be a big enough regular cardinal and let M H(θ) be such that K ∈ M. Then K/M is totally disconnected.~~

Proof Fix x,y ∈ K such that x ∼M y, and choose f : K → I in M such that f(x)< f(y). Find disjoint closed rational intervals I,J such that f(x)∈ I , f(y)∈ J . Then A = f −1[I] and B = f −1[J ] are disjoint closed subsets of K and A,B ∈ M.Using the zero-dimensionality of K and elementarity, ﬁnd a clopen set U ∈ M such that ⊆ ∩ = [ ] B U and A U . Then the characteristic function of U shows that x ∼M and [ ] y ∼M are separated by a continuous two-valued map. 17.2 Projections from elementary submodels 361

Proposition 17.2 There exists a continuous surjection ϕ : 2ω → I such that for every zero-dimensional compact space K the map T : C(K,2ω) → C(K,I) deﬁned by Tf = ϕ ◦ f is a surjection. In other words, for every g : K → I there is f : K → 2ω such that g = ϕ ◦ f .

~~Note that the map ϕ does not depend on K.~~

Proof First, observe that, given a zero-dimensional compact space K,everymap f : K → I factorizes through the Cantor set. More precisely, there are maps g : K → 2ω and h: 2ω → I such that f = h ◦ g. In order to see it, fix a countable ∈ = M ◦ M elementary submodel M of a big enough H(θ) that f M. Then f f qK and K/M is a zero-dimensional metric compact space, by Lemma 17.3. Thus K/M embeds into the Cantor set. Taking a retraction r : 2ω → K/M, we see that = M ◦ ◦ ◦ M f f r i qK , where i is the left-inverse to r. ω Thus, it suffices to prove our statement for K = 2 .LetZ be the Cech–Stoneˇ ω ×{ } compactification of the disjoint topological sum f ∈C(2ω,I) 2 f . Note that Z is totally disconnected. Let F : Z → I be the (uniquely defined) map satisfying F(x,f)= f(x)for (x, f ) ∈ 2ω ×{f }. Using the same factorization property as before, we obtain that F = G ◦ ϕ for some map G: Z → 2ω. It is clear that ϕ is the desired map.

~~17.2 Projections from elementary submodels~~

We show how to use elementary submodels for constructing projections in some Banach spaces. Let (X, ·) be a Banach space. Recall that a set D ⊆ X is norming if the formula (∗) |x|=sup{|ϕ(x)|/ϕ: ϕ ∈ D \{0}} (17.2) deﬁnes an equivalent norm on X. More precisely, we say that D is r-norming if x≤r|x| for every x ∈ X. D is 1-norming if |·|=·. The following fact is rather well known.

~~Proposition 17.3 Let D be a norming subset of X. Then D is 1-norming with respect to the norm |·|deﬁned by equation (17.2).~~

~~Proof Let D be the linear span of D, and let~~

~~ D1 ={ϕ ∈ D :ϕ≤1}. 362 17 Banach Spaces with Many Projections~~

~~|x|= |ϕ(x)| Then supϕ∈D1 . Thus, it remains to notice that D1 ={ϕ ∈ D :|ϕ|≤1}.~~

~~Indeed, if ϕ ∈ D1, then |ϕ(x)|≤1 whenever |x|≤1, so |ϕ|≤1. Conversely, if ϕ ∈ D \ D1, then there is x0 ∈ X such that |x0|=1 and |ϕ(x0)| > 1, so |ϕ| > 1.~~

~~The following lemma, in the case of WCG Banach spaces, was proved by Koszmider [245].~~

Lemma 17.4 Assume X is a Banach space, D ⊆ X is r-norming and M is an elementary substructure of a big enough (H (θ), ∈) that X, D ∈ M. Then ⊥ (a) XM ∩ (D ∩ M) ={0}; and ⊥ (b) the canonical projection P : XM ⊕ (D ∩ M) → XM has norm ≤ r.

Proof Fix x ∈ X ∩ M, y ∈ ⊥(D ∩ M) and ε>0. Since D is r-norming, there exists d ∈ D such that r|d(x)|/d≥x−ε. Since x ∈ M, by elementarity we may assume that d ∈ M. Thus d ∈ D ∩ M and d(y) = 0. It follows that

~~x≤r|d(x)|/d+ε = r|d(x + y)|/d+ε ≤ rx + y+ε.~~

⊥ By continuity, we see that x≤rx + y whenever x ∈ XM and y ∈ (D ∩ M). In particular, ⊥ XM ∩ (D ∩ M) ={0} ⊥ ⊥ because if x ∈ XM ∩ (D ∩ M), then we have −x ∈ (D ∩ M) and x≤rx − x=0.

⊥ Note that, in the lemma above, the subspace XM ⊕ (D ∩ M) is closed in X. ⊥ It may happen that (D ∩ M) ={0} (consider X = ∞), and in that case the lemma above is meaningless. We are going to discuss Banach spaces for which Lemma 17.4 provides a way to construct full projections. We demonstrate the use of elementary submodels for ﬁnding projections in WCG Banach spaces.

Proposition 17.4 Let X be a WCG Banach space, and let θ be a big enough regular cardinal. Furthermore, let M (H (θ), ∈) be such that X ∈ M. Then there exists a ⊥ norm-one projection PM : X → XM such that ker(PM ) = (X ∩ M).

Proof Let K be a linearly dense weakly compact subset of X. By Lemma 17.4,it ⊥ sufﬁces to check that XM ∪ (X ∩ M) is linearly dense in X. Suppose ϕ ∈ X \{0} is such that (X ∩ M) ⊆ ker(ϕ) and ⊥(X ∩ M) ⊆ ker(ϕ). The latter inclusion implies that ϕ ∈ (X ∩ M) (the closure in σ(X,X)) because 17.2 Projections from elementary submodels 363

X ∩ M is Q-linear. Fix p ∈ K such that ϕ(p) = 0. Let U0,U1 ⊆ R be disjoint open rational intervals such that 0 ∈ U0 and ϕ(p) ∈ U1.LetK0 be the weak closure of K ∩ M. Note that ϕ K0 = 0 because ϕ is weakly continuous. Using the fact that

~~ϕ ∈ (X ∩ M),~~

~~ for each x ∈ K0 choose ψx ∈ X ∩ M such that ψx(x) ∈ U0 and ψx(p) ∈ U1.By compactness, there are~~

~~x0,x1,...,xn−1 ∈ K0 such that (∗) K ⊆ ψ−1[U ]. 0 xi 0 (17.3) i~~

~~(∃ x ∈ K)(∀ ψ ∈ Ψ) ψ(x)∈ U1.~~

~~All parameters in the formula above are in M. Therefore, by elementarity, there ∈ ∩ ⊆ ∈ exists x K M K0 such that ψxi (x) U1 for each i~~

Fix a Banach space X and a norming set D ⊆ X. We shall say that D generates projections in X if, for a big enough regular cardinal θ, for every countable elementary substructure M (H (θ), ∈) with D ∈ M it holds that

~~⊥ X = XM ⊕ (D ∩ M).~~

We shall say that the projection P : X → X such that im P = XM and ker P = ⊥(D ∩ M) is induced by the triple (X,D,M). We ﬁnish this section with a short discussion of projectional generators. Let X be a Banach space. A pair (D, Φ) is a projectional generator in X if (1) D ⊆ X is a norming Q-linear subspace, ≤ℵ (2) Φ: D →[X] 0 , and ⊥ (3) ( Φ[B]) ∩ cl∗ B = 0 whenever B ⊆ D is Q-linear. Projectional generators were introduced by Orihuela and Valdivia [321]asatoolfor showing that certain Banach spaces have a projectional resolution of the identity. The ﬁrst projectional generators were implicitly constructed by Lindenstrauss [269, 270]. Refer to Chapter 6 of Fabian’s book [147] for more information. Below we show how projectional generators together with elementary submodels induce projections.

~~Proposition 17.5 Let X be a Banach space that has a projectional generator with domain D ⊆ X. Then D generates projections in X. 364 17 Banach Spaces with Many Projections~~

Proof Fix M (H (θ), ∈) with D ∈ M, where θ is big enough that H(θ) satisfies “there exists a projectional generator on X with domain D.” By elementarity, there is Φ ∈ M such that M satisfies “(D, Φ) is a projectional generator on X.” It suffices to ⊥ check that the only ψ ∈ X that vanishes on XM ∪ (D ∩ M) is the zero functional. Let B := D ∩ M. By elementarity, B is Q-linear because D is assumed to be Q-linear. By the definition of a projectional generator, ⊥ Φ[B] ∩ cl∗ B = 0. ⊥ ⊥ Thus, if ψ ∈ X is such that XM ∪ (D ∩ M) ⊆ ker ψ, then ψ ∈ ( Φ[B]) .Thisis because Φ(b) ⊆ M whenever b ∈ B (recall that Φ has countable values). It follows that ψ = 0.

~~17.3 Lindelöf property of weak topologies~~

It has already been mentioned that every WCG Banach space is weakly Lindelöf (even K-analytic) (see Section 12.4). We shall prove a general result on the Lindelöf property, using projections induced by elementary submodels.

~~Theorem 17.1 Let X be a Banach space, and let D ⊆ X be a norming space that generates projections in X. Then (X, σ (X, D)) is Lindelöf.~~

~~Proof Recall that a standard basis of the topology σ(X,D) consists of ﬁnite intersections of sets of the form~~

W ={x ∈ X : ϕ(x) < r}, where ϕ ∈ D and r is a rational number. Fix a cover V of X consisting of basic sets in σ(X,D). Let θ be a big enough regular cardinal, and let M (H (θ), ∈) be countable such that X, D ∈ M and V ∈ M. We claim that V ∩ M is a cover of X, which will of course complete the proof. Fix x ∈ X.LetP : X → X be the projection induced by (X,D,M).Findara- tional r>0 such that the ball B(Px,r) is contained in some element of V .Find z ∈ X ∩ M such that Px − z

In the case where X generates projections in X (equivalently, in the case where X is WLD, as we shall see later), the result above says that X is weakly Lindelöf. 17.4 Separable complementation property 365

~~17.4 Separable complementation property~~

We present selected results concerning the complementation property in general Ba- nach spaces in order to motivate the study of projectional skeletons and projectional resolutions. We start with the following classical result due to Sobczyk.

~~Theorem 17.2 (Sobczyk [384]) Every isomorphic copy of the sequence space c0 in a separable Banach space is complemented.~~

Proof Fix a separable Banach space X and ﬁx an embedding i : c0 → X so that ≤ −1≤ { } ∗ i r and i r.Let en n∈ω be the canonical basis of c0, and let en be the nth coordinate functional. Let K denote the closed unit ball of X. In what follows, we shall have in mind the weak∗ topology in X. Deﬁne

~~={ ∈ : = ∗} Fn y X i (y) en .~~

~~Observe that Fn ∩ rK = ∅ by the Hahn–Banach theorem. We prove the following.~~

~~Claim 17.1 Every neighborhood of 0 ∈ X intersects all but ﬁnitely many of the sets Fn ∩ 2rK.~~

~~Suppose otherwise and ﬁx a sequence {kn}n∈ω ⊆ ω and a neighborhood W of 0 ∩ =∅ ∈ { } so that W Fkn for n ω. Choose a sequence yn n∈ω such that ∈ ∩ \ yn Fkn rK W.~~

Taking a subsequence if necessary, we may assume that yn → y for some y ∈ rK (recall that we consider the weak∗ topology on rK, which is compact and metrizable). By continuity, i (y) = 0. Hence zn := yn − y is an element of Fn ∩ 2rK. Finally,

lim zn = 0, n→∞ and therefore ∈ ∩ ∩ zn Fkn 2rK W for all but ﬁnitely many n ∈ ω, a contradiction. Fix a metric on 2rK that induces the weak∗ topology. Given n ∈ ω, choose ϕn ∈ 2rK so that

~~(ϕn, 0)~~

Recall that a Banach space E has the separable complementation property if for every separable A ⊆ E there exists a projection P : E → E such that A ⊆ im P . 366 17 Banach Spaces with Many Projections

~~Corollary 17.3 Assume E is a Banach space having the separable complementation property. Then every copy of c0 is complemented in E.~~

Proof Assume c0 ⊆ E and ﬁx a projection P : E → E such that X = im P contains c0. By Sobczyk’s theorem, there exists a projection Q: X → X with im Q = c0. The composition Q ◦ P is a projection of E onto c0.

Let K denote the one-point compactification of the natural numbers (i.e., K = N ∪{∞}with the obvious topology). The space c0 is naturally identified with the subspace of C(K) consisting of all functions vanishing at ∞. More generally, if K is any compactification of N, then the subspace of C(K) consisting of functions that are constantly zero on K \ N is naturally isometric to c0. We say that this is a canonical copy of c0. The next theorem is a straight generalization of a result of Whitley [434]. The special case saying that c0 is not complemented in ∞ has already been proved by Sobczyk [384].

~~Theorem 17.3 Let K be a compactiﬁcation of the natural numbers N. If the canonical copy of c0 is complemented in C(K), then K \ N admits a strictly positive regular Borel measure.~~

~~Proof Set E ={f ∈ C(K): f (K \ N) = 0} and assume P : C(K) → E is a bounded projection. Deﬁne ϕn(f ) = P (f )(n), and let μn be a real-valued measure on K that induces ϕn,~~

ϕn(f ) = fdμn. K Deﬁne −n ν = 2 |μn|, n∈ω where |μn| denotes the variation of μn. Then ν is a positive regular Borel measure on K. We claim that ν is strictly positive on K \ N. Fix a nonempty open set U ⊆ K such that ν(U \ N) = 0 and ﬁx f ∈ C(K) with supp(f ) ⊆ U. Observe that |μn|(U \{n}) = 0, which implies that

~~P (f )(n) = fdμn = f(n) X for every n ∈ N. Thus f ∈ E (i.e., f(x)= 0 for every x ∈ K \ N). As f was chosen arbitrarily, it follows that U \ N =∅.~~

~~This yields the following corollary. 17.4 Separable complementation property 367~~

~~Corollary 17.4 (a) The space c0 is not complemented in ∞. (b) There exists a Banach space of density ℵ1 that has an uncomplemented copy of c0.~~

Proof (a) Recall that ∞ ≈ C(βN) and βN \ N does not satisfy the countable chain condition (and hence does not admit a strictly positive measure). (b) Let K be the one-point compactification of a locally compact space of the form N ∪ A , where A is an almost disjoint family of subsets of N and |A |=ℵ1. More precisely, A consists of infinite subsets of N, a ∩b is finite whenever a,b ∈ A are distinct, elements of N are isolated and a basic neighborhood of a ∈ A is of the form {a}∪(a \ s), where s is a finite set. Then K is a compactification of N whose remainder is homeomorphic to the one-point compactification of an uncountable discrete space. There is no strictly positive Borel measure on K \ N, and therefore the natural copy of c0 in C(K) is not complemented by Theorem 17.3.

~~Every quotient map f : K → L between compact spaces induces an isometric embedding f : C(L) → C(K), where f (ϕ) = ϕ ◦ f . A bounded linear operator~~

P : C(K) → C(L) satisfying P(ϕ ◦ f)= ϕ for ϕ ∈ C(L) is called an averaging operator for f .It is clear that its existence is equivalent to the fact that C(L) is complemented in C(K) when embedded via f . This happens, for example, when f is a topological retraction (i.e., if there is a continuous map j : L → K such that f ◦ j = id L). A classical result of Milyutin says that the standard quotient map f : 2ω → I, de- ﬁned by “gluing” all jumps in the Cantor set, does not admit any averaging operator. This gives an example of an uncomplemented copy of C(I) in a separable Banach space, C(2ω)—actually isomorphic to C(I). A more abstract result on the nonexistence of averaging operators for Cantor-type maps was proved by Pełczynski´ [326]. We present a similar result for increasing maps of compact lines. Its particular case is Milyutin’s theorem. A point x in a linearly ordered set X is internal if it is not isolated from either side (i.e., x = inf(x, →) and x = sup(←,x)). A linearly ordered compact space is called a compact line for short.

Theorem 17.4 (Kubis[´ 252]) Let θ : K → L be an increasing quotient of compact lines. If θ admits an averaging operator, then the set − Q ={y ∈ L:|θ 1(y)| > 1 and y is internal in L} is nowhere dense.

Proof For each x ∈ L, define x− = min θ −1(x) and x+ = max θ −1(x). Then each fiber of θ is of the form [x−,x+], where x ∈ L. Recall that θ identifies C(L) with 368 17 Banach Spaces with Many Projections the set of all f ∈ C(K) that are constant on every interval [x−,x+], where x ∈ L. Suppose P : C(K) → C(K) is a bounded linear projection onto C(L) embedded via θ in C(K).FixN ∈ ω such that −1 + N/3 ≥P .

Given p ∈ Q, choose an increasing function χp ∈ C(K) such that χp(t) = 0for − + t ≤ p and χp(t) = 1fort ≥ p .Lethp = Pχp. There exists a (unique) function hp ∈ C(L) such that hp = hpθ. Slightly abusing notation, we shall write hp instead of hp (i.e., we shall treat hp as a function on L.) Deﬁne − + Q ={q ∈ Q: hq (q) < 2/3},Q={q ∈ Q: hq (q) > 1/3}. Then at least one of the sets above is somewhere dense. Furthermore, deﬁne − = −1 −∞ + = −1 +∞ Up (hp) ( , 2/3), Up (hp) (1/3, ). − Suppose that the set Q is dense in the interval (a, b). Choose p0 pN−1. = − Let g f i

Now choose t ∈ U − ∩···∩U − p0 pN−1 such that t>pN−1. Note that |(Pf )(t)|≤P because 0 ≤ f ≤ 1. On the other hand, hpi (t) < 2/3forip1 > ···>pN−1 in Q+ so that p ∈ U + ∩···∩U + for i 1 (1/3) 1 N/3 P , i

A deeper analysis of the proof above can give a more precise result. This is done in [221]. In particular, the assumption that Q is nowhere dense is not sufﬁcient for the existence of an averaging operator. We refer to [221] for the details.

~~17.5 Projectional skeletons~~

In this section, we discuss Banach spaces with projectional skeletons, deﬁned below. Recall that a partially ordered set (poset for short) Γ is directed if for every s0,s1 ∈ Γ there is t ∈ Γ such that s0 ≤ t and s1 ≤ t. Γ is σ -complete if every sequence s0

(1) Each Ps is a projection of E. (2) Γ is an up-directed partially ordered set (i.e., for each s,t ∈ Γ , there is r ∈ Γ with s ≤ r and t ≤ r). 370 17 Banach Spaces with Many Projections

~~◦ = ◦ = ≤ (3) Ps Pt Pt Ps Ps whenever s t. = (4) E s∈Γ im Ps . ≤ ≤ = (5) If s0 s1 ... in Γ , then there exists t supn∈ω sn and = im Pt im Psn . n∈ω~~

~~So far we have not required that the projections in the skeleton s be uniformly bounded. We shall do so in view of the following fact.~~

~~Proposition 17.6 Let s ={Ps}s∈Γ be a projectional skeleton in a Banach space E. Then there exists a closed coﬁnal set Π ⊆ Γ such that~~

~~sup Pt < +∞. t∈Π~~

~~Proof For each n ≥ 1, deﬁne~~

~~Πn ={s ∈ Γ :Ps≤n}.~~

~~We first show that each Πn is closed in Γ . ≥ = Indeed, fix k 1, fix s0 k.Let = − ∈ ∈ ε (r k)/2. Using the second part of (5), find m ω and y im Psm such that~~

~~Pt (x − y) <ε/k.~~

~~= ◦ ≤ ≤ Note that Psm Psm Pt . Using the fact that Psm k, we get Psm (x) k and − = − ≤ − y Psm (x) Psm (y Pt (x)) k y Pt (x) <ε.~~

~~Thus~~

~~ ≤ − + − + + + ≤ + = = Pt x Pt x y y Psm x Psm x <ε/k ε k 2ε k r Pt x .~~

This contradiction shows that Pt ≤k. Finally, we claim that one of the sets Πn is coﬁnal in Γ . Suppose otherwise, and for each n ∈ ω choose tn such that Ps >nwhenever tn ≤ s. Using the directedness of Γ , we construct a sequence s1

~~From now on, we shall require that a projectional skeleton s ={Ps}s∈Γ satisfy +∞ (6) sups∈Γ Ps < .~~

~~We shall say that s is a k-projectional skeleton if Ps≤k for each s ∈ Γ . 17.5 Projectional skeletons 371~~

~~Proposition 17.7 Let s ={Ps}s∈Γ be a projectional skeleton in a Banach space E, and let x ∈ E. Then~~

~~Pt x = lim Ps x n→∞ n ≤ ≤ = whenever s0 s1 ... in Γ is such that t supn∈ω sn.~~

~~= ∈ Proof Let r sups∈Γ Ps , and ﬁx ε>0. Using the second part of (5), ﬁnd y n∈ω im Psn such that~~

~~Pt x − y <ε/(1 + r). ∈ = = ≥ Choose k such that y im Psk . Note that Pt y y and Psn y y for n k. Thus, given n ≥ k,wehave~~

~~ − ≤ − + − + + − Pt x Psn x Pt x y y Psn x <ε/(1 r) Psn (y Pt x)~~

~~≤ ε/(1 + r)+ ry − Pt x <ε/(1 + r)+ rε/(1 + r)= ε. − = This shows that limn→∞ Pt x Psn x 0.~~

~~The fact above provides a simple way to construct projections onto nonseparable subspaces.~~

~~Proposition 17.8 Let s ={Ps}s∈Γ be a projectional skeleton in a Banach space E, and let T ⊆ Γ be a directed set. Then the formula~~

~~PT x = lim Psx s∈T well deﬁnes a bounded projection of E onto ( s∈T im Ps).~~

Proof It is enough to show that {Psx}s∈T is a Cauchy net for every x ∈ E. Suppose this is not the case, and ﬁx x ∈ E and ε>0 such that for each s ∈ T there are ∈ ≤ ≤ − t1,t2 T , s t1, s t2 with Pt1 x Pt2 x > 2ε.AsT is directed, we have that for each s ∈ T there are t1,t2 ∈ T with s ≤ t1 ≤ t2 and − Pt1 x Pt2 x >ε.

~~Hence, by induction, we can construct a sequence~~

~~t1 ≤ t2 ≤ t3 ≤ t4 ≤ ... in T such that − Pt2k−1 x Pt2k x >ε for k ∈ N. This contradicts Proposition 17.7. 372 17 Banach Spaces with Many Projections~~

Below we recall the notion of a projectional resolution of the identity, a useful tool for investigating some classes of nonseparable Banach spaces. The idea of “long sequences of projections” goes back to Lindenstrauss [269, 270]. For more information and historical comments, we refer to the books [107] and [147]. Let X be a Banach space, and let λ be a limit ordinal. A projectional sequence of length λ in X is a sequence of projections {Pξ }ξ<λ satisfying the following conditions: ⇒ = ◦ = ◦ (i) ξ<η Pξ Pη Pξ Pξ Pη. (ii) P X = ( P X) for every limit ordinal δ<λ. δ ξ<δ ξ = (iii) X ( ξ<λ Pξ X). A special case is a projectional resolution of the identity (PRI). This is a projectional sequence {Pξ }ξ<λ such that Pξ =1 and dens (Pξ X) ≤|ξ|+ℵ0 for each ξ<λ, where |ξ| denotes the cardinality of the ordinal ξ. The existence of a reasonable projectional sequence or PRI in a given Banach space allows proving certain properties by transﬁnite induction. On the other hand, some Banach spaces can be built from smaller spaces using projections of norm one. This is described in the following theorem.

Theorem 17.5 (Kubis[´ 252]) Assume E is a Banach space that is the union of a continuous chain {Eα}α<κ of closed subspaces such that, for every α<κ, Eα is 1-complemented in Eα+1. Then there exist projections Sα : E → E, α<κ, such that Sα=1, imSα = Eα and Sα ◦ Sβ = Sα = Sβ ◦ Sα whenever α ≤ β<κ.

β Proof We inductively construct projections {Sα : α ≤ β ≤ κ} with the following properties: β (a) Sα : Eβ → Eα has norm one for every α ≤ β, β ◦ γ = γ ≤ ≤ (b) Sα Sβ Sα whenever α β γ . 0 = ≤ β We start with S0 id E0 .Fix0<δ κ, and assume Sα has been constructed for each α ≤ β<δand satisﬁes conditions (a) and (b). There are two cases: Case 1. δ = γ + 1. Using the assumption, ﬁx a projection

~~T : Eγ +1 → Eγ = δ = γ ◦ with T 1, and deﬁne Sα Sα T for every α<δ. Clearly, both (a) and (b) are satisﬁed. Case 2. δ is a limit ordinal. Let G = Eα. α<δ~~

~~Then G is a dense linear subspace of Eδ. Deﬁne hα : G → Eα by setting = β = { ∈[ : ∈ } hα(x) Sα x, β min ξ α, δ) x Eξ . 17.5 Projectional skeletons 373~~

Note that if α<β, x ∈ Eβ and γ ∈[β,δ), then γ = β γ = β ∗ Sα x Sα (Sβ x) Sα x ( ) γ = ∗ because of (b) and by the fact that Sβ Eβ id Eβ .Using( ), it is easy to see that hα is a linear operator. Clearly, hα=1, and thus it can be uniquely extended to a linear operator δ : → Sα Eδ Eα. Finally, δ = = Sα hα 1 δ and Sα is a projection onto Eα. Thus (a) holds. It remains to show (b). Fix α<β<δ. By continuity, it suffices to check that δ = β δ ∈ ∈ ∈[ Sαx Sα Sβ x holds for every x G.Fixx G, and find γ β,γ) such that x ∈ Eγ .Wehave β δ = β γ = γ = δ Sα Sβ x Sα Sβ x Sα x Sαx. Thus conditions (a) and (b) hold. It follows that the construction can be carried out. := κ = Finally, define Sα Sα . Clearly, Sα is a projection of E onto Eα and Sα 1. If α<β<κ, then = κ ◦ κ = β ◦ κ ◦ κ = β ◦ κ = κ = SαSβ Sα Sβ Sα Sβ Sβ Sα Sβ Sα Sα, and of course Sβ Sα = Sα, because Eα ⊆ Eβ . This completes the proof.

~~Theorem 17.6 (Kubis[´ 250]) Every Banach space with a 1-projectional skeleton has a projectional resolution of the identity onto subspaces with the same property.~~

Proof Fix a Banach space E with κ = dens E, and let s ={Ps}s∈Γ be a projectional skeleton in E such that Ps=1 for every s ∈ Γ . Furthermore, ﬁx a continuous chain {Tα}α<κ of up-directed subsets of Γ satisfying |Tα|≤α +ℵ0 for each α and such that {PsE : s ∈ Tα,α<κ} is dense in E. Continuity of the chain means that T = T whenever δ is a δ ξ<δ ξ limit ordinal. Let X = ( P E). Then {X } is a chain of closed subspaces α s∈Tα s α α<κ of E, the density of Xα does not exceed |α|+ℵ0, and Xδ = Xξ ξ<δ for every limit ordinal δ. By Proposition 17.8,formula

Pαx = lim Psx s∈Tα 374 17 Banach Spaces with Many Projections deﬁnes a projection of E onto Xα. Clearly, Pα=1 because Ps=1fors ∈ Γ . Fix α<β. Then Tα ⊆ Tβ , and therefore Pβ ◦ Pαx = Pα. Observe that Ps ◦ Pβ = Ps for every s ∈ Tα. Indeed, given s ∈ Tα,theset

~~A ={t ∈ Tβ : t ≥ s}⊆Tβ is coﬁnal in Tβ ,so~~

~~PsPβ x = lim PsPt x = lim PsPt x = Psx. t∈Tβ t∈A~~

~~It follows that~~

~~PαPβ x = lim PsPβ x = lim Psx = Pαx. s∈Tα s∈Tα~~

~~Thus, {Pα}α<κ is a PRI on E. Finally, for each α,letΓα be the smallest σ -closed subset of Γ that contains Tα. { } Then Ps Xα s∈Γα is a projectional skeleton in Xα.~~

Corollary 17.5 Let E be a Banach space that has a projectional skeleton. Then E has a locally uniformly convex equivalent norm and admits a bounded injective linear operator into c0(S) for some set S.

Proof The ﬁrst statement is a well-known result of Zizler [437]; we omit its rather technical proof. The second statement is also well known and can be proved by induction. We sketch the argument here. Assume {Pα}α<κ is a PRI on E such that P0 = 0 and each space Xα = im Pα has an injective operator

~~Qα : Xα+1 → c0(Sα) with Qα≤1. Let Y be the c0-sum of the family~~

~~{c0(Sα)}α<κ. = = Then Y c0(S), where S α<κ Sα, assuming that the sets Sα are pairwise disjoint. Deﬁne Q: E → Y by~~

~~(Qx)(α) = QαPα+1x − QαPαx, x ∈ E.~~

~~We need to check that Q is well deﬁned. Fix x ∈ E and ε>0. We claim that~~

Pα+1x − Pαx <ε for all but ﬁnitely many α<κ. Once we prove it, using the fact that Qα≤1, we conclude that Qx ∈ Y . Suppose that − ≥ Pαn+1x Pαn x ε 17.6 Norming subspaces induced by a projectional skeleton 375 for some sequence {αn}n∈ω. We may assume that this sequence is strictly increasing. = Let β supn∈ω αn. Then

Pβ x = lim Pξ x, ξ<β and hence there is η<βsuch that Pξ x − Pαx <εfor every η<ξ,α<β.This is a contradiction. It is clear that Q ≤ 1. To see that ker Q = 0, ﬁx x ∈ E \{0} and let be the minimal ordinal ξ<κfor which Pξ x = 0. Then = α + 1 and by assumption QαPα+1x = 0, so (Qx)(α) = 0.

~~17.6 Norming subspaces induced by a projectional skeleton~~

~~We shall now look at the dual of a space with a projectional skeleton. Let s = {Ps}s∈Γ be a projectional skeleton in a Banach space X.Theset = ∗ D Ps X s∈Γ~~

~~∗ ∩ is clearly a linear subspace of X. Notice that Ps X BX endowed with the topology ∗ σ(X ,X) is second-countable because Ps X is linearly homeomorphic to the dual of PsX.Let~~

r = sup Ps. s∈Γ Given x ∈ SX, there is s ∈ Γ such that x = Psx; choose ϕ ∈ X such that ϕ(x) = 1 =ϕ. Then ∗ = = = (Ps ϕ)x ϕ(Psx) ϕ(x) 1 and ∗ ≤ Ps ϕ r. This shows that the space D is r-norming. We shall say that D is the dual norming subspace induced by s and shall denote it by D(s). Let s ={Ps}s∈Γ and t ={Qt }t∈Π be projectional skeletons in the same Banach space X. We say that s andt are equivalent if they induce the same norming sub- ∗ = ∗ space (i.e., s∈Γ im Ps t∈Π im Qt ). It turns out that, with the help of elementary submodels, a projectional skeleton can be recovered (up to equivalence) from the norming space.

Lemma 17.5 Let s ={Ps}s∈Γ be a projectional skeleton in a Banach space X, and let D ⊆ D(s) be norming for X. Furthermore, let θ be a big enough regular cardinal and let M (H (θ), ∈) be countable and such that s ∈ M. Then the projection induced by (X,D,M)equals Pt , where t = sup(Γ ∩ M). 376 17 Banach Spaces with Many Projections

Proof Observe that im Ps ⊆ (X ∩ M). s∈Γ ∩M This is because, given s ∈ Γ ∩M, by elementarity there exists a countable set A ∈ M that is dense in im Ps . By Proposition 17.1(c), A ⊆ X ∩ M, and therefore im Ps ⊆ (X ∩ M). It follows that im Pt = im Ps ⊆ (X ∩ M). s∈Γ ∩M On the other hand, given x ∈ X ∩ M, by elementarity there is s ∈ Γ ∩ M such that x ∈ im Ps ; thus x ∈ im Pt . Hence im Pt = XM . Notice that, again by elementarity, = Pt ϕ ϕ whenever ϕ ∈ D ∩ M. Thus ⊥ ker Pt ⊆ (D ∩ M) because if Pt x = 0 and ϕ ∈ D ∩ M, then = = = ϕ(x) (Pt ϕ)x ϕ(Pt x) 0. ⊥ It follows that ker Pt = (D ∩ M) because X = XM ⊕ ker Pt and ⊥ XM ∩ (D ∩ M) ={0} by (Lemma 17.4(a)). This completes the proof.

Theorem 17.7 (Kubis[´ 250]) Let X be a Banach space, and let D ⊆ X be r- norming (r ≥ 1). The following properties are equivalent: (a) X has an r-projectional skeleton s such that D ⊆ D(s). (b) D generates projections in X.

Proof Implication (a) ⇒ (b) follows from Lemma 17.5. (b) ⇒ (a). Fix a big enough regular cardinal θ, and let Γ be the family of all countable elementary substructures M of (H (θ), ∈) such that D ∈ M. Endow Γ with inclusion. Clearly, Γ is a σ -directed poset. Fix M ∈ Γ , and let PM be the projection onto (X ∩ M) with ⊥ ker(PM ) = (D ∩ M).

By deﬁnition, PM is deﬁned on the entire space. Furthermore, PM ≤ r by ⊆ ⊆ = Lemma 17.4. Given a sequence M0 M1 ... in Γ , the union M n∈ω Mn is an elementary substructure of H(θ) such that (X ∩ M) is the closure of (X ∩ Mn). n∈ω 17.6 Norming subspaces induced by a projectional skeleton 377

~~It follows that~~

~~s ={PM }M∈Γ ⊆ ∩ ⊆ is a projectional skeleton in X. It is clear that D D(s) because D M im PM .~~

~~We note the following corollary.~~

~~Corollary 17.6 Let X be a Banach space with a projectional skeleton. Then there exists a renorming of X under which X has an equivalent 1-projectional skeleton.~~

Proof Let D be the dual norming subspace induced by a ﬁxed projectional skeleton in X. By Lemma 17.5, D generates projections in X. Consider a renorming of X after which D becomes 1-norming (see Proposition 17.3). By Theorem 17.7, X has a 1-projectional skeleton.

~~As an application of Theorem 17.7, we prove that the class of Banach spaces with a 1-projectional skeleton is stable under arbitrary c0- and p-sums.~~

Theorem 17.8 (Kubis[´ 250]) Let {Xα}α<κ be a collection of Banach spaces, and let = ≤ ∞ X α<κ Xα be endowed either with the c0-norm or the p-norm (1 p< ). Furthermore, assume that, for each α<κ, Dα ⊆ Xα is 1-norming and generates projections in Xα. Then the set D ={ϕ ∈ X : (∀ α) ϕ Xα ∈ Dα and |{α : ϕ Xα = 0}| ≤ ℵ0} is 1-norming and generates projections in X.

~~Proof The fact that D is 1-norming follows from the properties of the c0-sum and the p-sum. Deﬁne~~

supp(ϕ) ={α : ϕ Xα = 0}. Fix a suitable M (H (θ), ∈) so that D ∈ M.LetS = κ ∩M. Note that supp (ϕ) ⊆ S whenever ϕ ∈ D ∩ M. Suppose ⊥ X = XM ⊕ (D ∩ M), and ﬁx ψ ∈ X satisfying X ∩ M ⊆ ker ψ and ⊥ (D ∩ M) ⊆ ker ψ.

~~Then ψ is in the weak∗ closure of the linear hull of D∩M, and therefore supp (ψ) ⊆ S. Assuming ψ = 0, there is α ∈ S such that~~

~~ψα := ψ Xα = 0. ⊥ Note that Xα ∩ M ⊆ ker ψα.Ifx ∈ (Dα ∩ M) and ϕ ∈ D ∩ M, then~~

~~ϕ Xα ∈ Dα ∩ M 378 17 Banach Spaces with Many Projections so ϕ(x) = 0. It follows that~~

~~⊥ ⊥ (D ∩ M) ∩ Xα = (Dα ∩ M).~~

~~⊥ Thus (Dα ∩ M) ⊆ ker ψ. On the other hand,~~

~~⊥ Xα = (Xα ∩ M)⊕ (Dα ∩ M) because Dα ∈ M. This is a contradiction.~~

Given a Banach space E, it is natural to ask when E generates projections in E as a subspace of E. It turns out that this happens exactly in the case where E is Asplund. This is a result of Fabian and Godefroy [148]. Recall that a Banach space E is Asplund if the dual of every separable subspace of E is separable. We have the following theorem.

~~Theorem 17.9 Let E be a Banach space. The following properties are equivalent: (a) E is an Asplund space. (b) E ⊆ E generates projections in E.~~

Proof That (a) =⇒ (b) is a consequence of the existence of a projectional generator in E with domain E. This is a highly nontrivial result of [148]. For details, we refer either to [148] or to Fabian’s book [147]. Below we justify the implication (b) =⇒ (a). Assume E generates projections in E, and ﬁx a separable subspace Y of E.Fixa countable M H(θ)such that E ∈ M and Y ∩M is dense in Y , and let P : E → E be the projection with im P = (E ∩ M) and ker P = (E ∩ M)⊥. Then P y = y for every y ∈ Y because Y is contained in the weak∗ closure of E ∩ M.Fixϕ ∈ Y , and let ψ ∈ E be an extension of ϕ. Then

~~(P ψ)y = ψ(Py) = ψ(y)= ϕ(y) for every y ∈ Y . Thus, ϕ = (P ψ) Y . It follows that Y is separable because so is im P .~~

~~Next we exhibit a topological property of norming spaces induced by a projectional skeleton.~~

Theorem 17.10 (Kubis[´ 250]) Let X be a Banach space with a projectional skeleton s ={Ps}s∈Γ , and let D ⊆ X be the norming space induced by s, endowed with the weak∗ topology. Then: (a) The weak∗closure in X of the convex hull of every countable bounded subset of D is metrizable and contained in D. (b) D has countable tightness. 17.6 Norming subspaces induced by a projectional skeleton 379

= Proof Part (a) is trivial: Every countable subset of D is contained in Ys Ps X for ∗ some s ∈ Γ , and every bounded subset of Ys with the weak topology is second- countable. (b) Let A ⊆ D and p ∈ (A) ∩ D be given, where the closure is taken in σ(X,X). Replacing A by A−p, we may assume that p = 0. Fix a big enough regular cardinal θ and a countable elementary substructure M of (H (θ), ∈) such that X, s,A∈ M. We claim that 0 ∈ (A ∩ M). = ∩ = ∗ Let t sup(Γ M), and let Yt Pt X .Fixaweak neighborhood U of p.We may assume that U = U(xi,ε), i

x0,...,xn−1 ∈ X, and ε>0 is rational. By Lemma 17.5, Pt X = (X ∩ M). By Banach’s open mapping −1 principle, Pt [X ∩ M] is dense in X. Hence, without loss of generality, we may assume that Pt xi ∈ M for each i

~~Thus a ∈ U(xi,ε). Finally, a ∈ A ∩ M ∩ U.~~

~~Corollary 17.7 Let X be a Banach space, and let D,E ⊆ X be norming spaces induced by projectional skeletons. If D ∩ E is total, then D = E.~~

Proof Note that D ∩ E is a linear space. Assuming it is total, it must be weak∗ dense. Thus, given p ∈ D, we have that p ∈ (D ∩ E),sop ∈ (A) for some countable A ⊆ D ∩ E (Theorem 17.10(b)). By Theorem 17.10(a), we have

~~(A) ⊆ D ∩ E.~~

~~This shows that D ⊆ E. By symmetry, D = E.~~

Corollary 17.8 Let X be a Banach space, and let S ⊆ X generate projections in X. Furthermore, let D ⊇ S be the smallest linear subspace of X such that (A) ⊆ D for every countable set A ⊆ D, where the closure is taken in σ(X,X). Then D is induced by a projectional skeleton in X. 380 17 Banach Spaces with Many Projections

~~Proof By Theorem 17.7, there exists a projectional skeleton s in X such that S ⊆ D(s). By Theorem 17.10(a), we have~~

~~D ⊆ D(s).~~

On the other hand, D is weak∗ dense because it is a norming (and hence total) linear space. Hence, given p ∈ D(s), we have that p ∈ (D) so, by Theorem 17.10(b), one gets p ∈ (A) for some countable A ⊆ D. Hence p ∈ D. This shows that D = D(s).

We ﬁnish this section characterizing the weak Lindelöf property for Ba- nach spaces with projectional skeletons. A version involving projectional generators (instead of projectional skeletons) was proved by Cascales, Orihuela and Namioka [91].

Theorem 17.11 Let X be a Banach space with a projectional skeleton. Then the following properties are equivalent: (a) X is weakly Lindelöf. (b) X has property (C) of Corson. (c) X generates projections in X.

Proof (a) =⇒ (b) is trivial. (b) =⇒ (c): Assume D ⊆ X is a norming subspace induced by a ﬁxed projectional skeleton in X. We need to show that D = X. By Theorem 17.10(a), the weak∗ closed linear span of every countable subset of D is contained in D. Since X has property (C), by Theorem 13.2 X has countable convex tightness (i.e., given y ∈ D¯ ∗, there exists a countable A ⊆ D such that y belongs to the weak∗ closure of the convex hull of A). It follows that D = X and hence X generates projections in X. (c) =⇒ (a): If X generates projections in X, then X is σ(X,X)-Lindelöf by Theorem 17.1.

~~We have already mentioned that a Banach space is WLD if and only if its dual generates projections. Since this is not an original deﬁnition, we shall explain the details later in Section 19.12.~~

~~17.7 Sigma-products~~

Below we present several properties of Σ(A)-products that will be needed later. Some of them are interesting on their own. The following notion is a special case of the one from (9.2). 17.7 Sigma-products 381

~~Recall again that a Σ(A)-product is a topological space of the form~~

A Σ(A)={x ∈ R :|supp(x)|≤ℵ0}, where supp(x) ={α : x(α) = 0} denotes the support of x ∈ RA. There is a more general notion of a Σ-product of general topological spaces; however, we shall concentrate on Σ-products of real lines or even closed real intervals since this is sufﬁcient for our purposes. The following property is well known.

~~Proposition 17.9 Σ(κ) has countable tightness for every κ.~~

~~Proof Fix p ∈ A ⊂ Σ(κ) and ﬁx a countable M H(χ) so that p,A ∈ M and χ is big enough. Let B = A ∩ M. We claim that p ∈ B. Fix a standard neighborhood V of p,~~

~~V ={x : x(i) ∈ Ii for i ∈ s}, where s ⊂ κ is ﬁnite and each Ii is an open rational interval. Let S = κ ∩ M, and let~~

~~W ={x : x(i) ∈ Ii for i ∈ s ∩ S}.~~

Then W is a neighborhood of p and W ∈ M. Observe that supp(p) ⊆ S,sop(i) = 0 for i ∈ S \ s. Since M |= W ∩ A = ∅, by elementarity, there exists a ∈ A ∩ M = B such that a ∈ W . Then

~~supp(a) ⊆ M ∩ κ = S and hence a(i) = 0 = p(i) ∈ Ii for i ∈ S \ s. It follows that a ∈ V . Thus, ﬁnally a ∈ B ∩ V .~~

The next, rather technical, statement will be needed later. It implies in particular that Σ-products are normal topological spaces and that [0, 1]κ is the Cech–Stoneˇ compactiﬁcation of Σ(κ) ∩[0, 1]κ . These facts were proved by Corson [104] and Glicksberg [182], respectively. κ Given x ∈ R and S ⊆ κ, we shall denote by x | S the function x · χS (i.e., (x | S)(α) = x(α) for α ∈ S and (x | S)(α) = 0forα/∈ S).

Lemma 17.6 Let X ⊆[0, 1]κ be such that X ∩ Σ(κ) is dense in X, and let f : X → Y be a continuous map. Let B be an open base for f [X]. Assume χ is big enough and M H(χ) is such that f,κ ∈ M and B ⊆ M. Let S = M ∩ κ. Then: (a) x | S ∈ (X) for every x ∈ X, where the closure is taken in [0, 1]κ . (b) If x ∈ X and x | S ∈ X, then f(x)= f(x| S). 382 17 Banach Spaces with Many Projections

Proof (a) Fix x ∈ X and fix a standard basic neighborhood u of x | S. That is, u is defined by a finite function, say ψ, whose values are open rational intervals. Let u denote the neighborhood defined by the function ψ restricted to S. Then u ∈ M. Now M |= u ∩ X = ∅, so there exists y ∈ M with y ∈ Σ(κ)∩ u ∩ X. Thus supp(y) ⊆ M (because it is countable) and therefore supp(y) ⊆ S. It follows that y ∈ u,so u ∩ X = ∅.Asu is arbitrary, we get x | S ∈ X. (b) Suppose x,x | S ∈ X and f(x)= f(x| S).Letv,w ∈ B be disjoint and such that f(x)∈ v, f(x| S) ∈ w. There exist basic neighborhoods u1,u2 of x and x | S, respectively, such that

f [u1]⊆v and f [u2]⊆w. Assume each ui is defined by a finite function ψi whose values are open rational intervals. Shrinking both neighborhoods if necessary, we may assume that ψ = ψ1 S = ψ2 S.Letu denote the basic open set defined by ψ. Then u ∈ M and M |= f −1[v]∩u = ∅. Thus, there exists y ∈ M such that

~~y ∈ Σ(κ)∩ X ∩ u and f(y)∈ v. But supp(y) ⊆ M; hence y(i) = 0fori/∈ S and therefore y ∈ u2 because~~

~~(x | S)(i) = 0 ∈ ψ2(i) for i/∈ S. It follows that f(y)∈ w, a contradiction.~~

~~Corollary 17.9 Let A and B be disjoint closed subsets of Σ(κ). Then (A) ∩ (B) = ∅, where the closure is taken in Rκ . In particular, Σ(κ) is a normal topological space.~~

~~Proof First, replace R by the open interval (0, 1), so that~~

~~A,B ⊆[0, 1]κ .~~

Suppose there is p ∈ (A) ∩ (B) in [0, 1]κ , and ﬁx a countable M H(χ) so that A,B,p ∈ M.LetS = κ ∩ M, and observe that supp(p) ⊆ S because p ∈ M. By Lemma 17.6(a), we have that 17.8 Markushevich bases, Plichko spaces and Plichko pairs 383

~~p = p | S ∈ (A) ∩ (B). Hence p ∈ A ∩ B, a contradiction.~~

~~17.8 Markushevich bases, Plichko spaces and Plichko pairs~~

~~A biorthogonal system in a Banach space E is a sequence of pairs~~

~~{(xα,yα)}α∈S such that~~

~~{xα : α ∈ S}⊆E,~~

{yα : α ∈ S}⊆E , and yα(xα) = 1 and yα(xβ ) = 0 whenever α = β. If, moreover, the set {xα : α ∈ S} is linearly dense in E, the system is called complete. A Markushevich basis in E is a complete biorthogonal system

~~{(xα,yα)}α∈S { : ∈ } ={ } such that the set yα α S is total for E (i.e., α∈S ker yα 0 ). A Markushe- vich basis~~

~~{(xα,yα)}α∈S is norming if the space~~

~~ D<ω ={y ∈ E :|{α : y(xα) = 0}| < ℵ0} is norming for E. If the space~~

Dω ={y ∈ E :|{α : y(xα) = 0}|≤ℵ0} is norming for E, we say that {(xα,yα)}α∈S is a countably norming Markushevich basis. Note that the space D<ω is the linear span of {yα : α ∈ S}, while the space Dω ∗ is the countably weak closed linear span of {yα : α ∈ S} (i.e., Dω is the smallest linear space V ⊆ E that contains {yα : α ∈ S} and has the property A ⊆ V whenever A ⊆ V is countable and the closure is taken in σ(E,E)). In the deﬁnitions above, we shall often say “k-norming” or “countably k-norming” if the respective spaces are k-norming. The following result is due to Markushevich. We refer readers to a recent book of Hajek et al. [198] for more information on biorthogonal systems in Banach spaces and related subjects. 384 17 Banach Spaces with Many Projections

~~Theorem 17.12 Every separable Banach space has a 1-norming Markushevich basis.~~

Proof Let X be a separable Banach space. Let {xn}n∈ω enumerate a ﬁxed, dense subset of X, and let {yn}n∈ω be a sequence of norm-one functionals such that yn(xn) =xn. Then the linear span of {yn}n∈ω is 1-norming for X. We construct by induction a biorthogonal system

~~{(en,fn)}n∈ω so that~~

~~span{xi}i~~

~~span{yi}i~~

Notice that fi(ei) = δi,j for i, j < 2. ∈ { } + At step 2n, choose xk2n / span ei i<2n and construct e2n ﬁrst, while at step 2n 1 choose ∈ { } y2n+1 / span fi i<2n+1 and construct f2n+1 ﬁrst. Finally, we obtain a Markushevich basis

~~{(en,fn)}n∈ω that is 1-norming because the linear span of {fn}n∈ω contains {yn}n∈ω.~~

We shall see later that many of the well-known nonseparable Banach spaces have a countably norming Markushevich basis. Below we give an example that distinguishes the properties of having norming and countably norming Markushevich bases. The next statement follows from other results; however, we give a direct proof here.

~~Proposition 17.10 The space C(ω1 + 1) has a countably 1-norming Markushevich basis. 17.8 Markushevich bases, Plichko spaces and Plichko pairs 385~~

~~Proof For each α ≤ ω1,letxα be the characteristic function of the interval [0,α]. Given α<ω1,letyα be the functional corresponding to the measure~~

δα − δα+1, ∈ + where δx denotes the Dirac measure of x ω1 1. Finally, let yω1 be the functional corresponding to the Dirac measure of 0 ∈ ω1 + 1. It is clear that { } (xα,yα) α∈ω1+1 is a biorthogonal system in C(ω1 + 1). Fix μ ∈ C(ω1 + 1) \{0}, = = = · and let K be the support of μ.Letα min K.Ifα ω1, then μ r δω1 , where r = 0; hence = = μ(xω1 ) r 0.

If α<ω1, then μ(xα) = μ({α}) = 0. It follows that { } xα α≤ω1 is linearly dense. { } Let D denote the countable closure of the linear span of yα α≤ω1 . We claim that D contains the 1-norming set

~~S ={δξ : ξ<ω1}.~~

~~Suppose S ⊆ D, and let α be minimal such that δα ∈/ D. Then α>0 because = ∈ δ0 yω1 D.~~

~~Furthermore, α cannot be a limit ordinal because, given ξ0 <ξ1 <ξ2 <... with = α supn ξn, we have that~~

~~δα = lim δξ n→∞ n in the weak∗ topology. Thus α = η + 1. But~~

~~δα = δη − (δη − δη+1) = δη − yη ∈ D, a contradiction.~~

It is remarkable that the space C(ω1 + 1) does not have a norming Markushevich basis. This is a result of Alexandrov and Plichko [2]. Another interesting result, due to Kalenda, says that C(ω2 + 1) does not have a countably norming Markushevich basis; see [217]. In fact, Kalenda’s result is valid for every regular cardinal κ>ω1. On the other hand, it is unknown whether it holds for arbitrary ordinals, namely 386 17 Banach Spaces with Many Projections whether the existence of a countably norming Markushevich basis on C(η + 1) implies that η<ω2. We now turn to the crucial notion of this section. A Banach space E is called a Plichko space if there are a linearly dense set A ⊆ E and a norming set B ⊆ E such that

|{x ∈ A: y(x) = 0}|≤ℵ0 for every y ∈ B. In that case, (E, B) will be called a Plichko pair and A will be called a witness for this property. We shall add a prefix “k-” in the definitions above whenever k ≥ 1 is such that B is k-norming. Note that if (E, B) is a Plichko pair witnessed by A ⊆ E, then the formula T (y)(a) = y(a) defines a one-to-one weak∗ continuous linear operator

T : E → RA such that B ⊆ T −1[Σ(A)]. Conversely, given a one-to-one weak∗ continuous linear operator as above, then (E, T −1[Σ(A)]) is a Plichko pair. The subspace T −1[Σ(A)] will sometimes be called a Σ-space or, more precisely, a Σ-subspace of E.

Proposition 17.11 Let E be a Banach space, and let k ≥ 1. The following properties are equivalent: (a) E is a k-Plichko space. (b) There exists a weak∗ continuous injective linear operator T : E → RΓ such that T −1[Σ(Γ)] is k-norming.

~~Proof (a) ⇒ (b): Let (E, B) be a k-Plichko pair with witness A, and deﬁne~~

T : E → RA by Tf = f A. By deﬁnition, Tf ∈ Σ(A) whenever f ∈ B. ∗ (b) ⇒ (a): Since T is weak continuous, for each γ ∈ Γ there is xγ ∈ E such Γ that prγ T = xγ , where prγ : R → R is the projection onto the γ th coordinate. Let

~~A ={xγ : γ ∈ Γ }, and let B = T −1[Σ(Γ)]. Then A witnesses that (E, B) is a k-Plichko pair.~~

~~Using PRIs, we shall prove that Plichko spaces are precisely those Banach spaces that have a countably norming Markushevich basis. 17.8 Markushevich bases, Plichko spaces and Plichko pairs 387~~

~~Theorem 17.13 Let (E, D) be a Plichko pair. Then D generates projections in E.~~

~~Proof Fix M (H (θ), ∈) such that D ∈ M. Suppose ϕ ∈ E is such that E ∩ M ⊆ ker ϕ and~~

~~⊥ (D ∩ M) ⊆ ker ϕ.~~

The latter fact means that ϕ ∈ (D ∩ M), where the closure is taken in σ(E,E).LetG ⊆ E be a linearly dense set such that supp(y, G) is countable for each y ∈ D. By elementarity, we may assume that G ∈ M. Suppose ϕ = 0, and ﬁx u ∈ G such that ϕ(u) = 0. Since ϕ is in the weak∗ closure of D ∩ M, we may ﬁnd ψ ∈ D ∩ M such that ψ(u)= 0. Thus

~~u ∈ supp(ψ, G).~~

~~On the other hand, supp(ψ, G) ∈ M because ψ,G ∈ M. By Proposition 17.1(c), one gets supp(ψ, G) ⊆ M. In particular, u ∈ E ∩ M and hence ϕ(u) = 0, a contradiction.~~

A similar statement was proved by Koszmider [245, Lemma 4.1] using the existence of a countably 1-norming Markushevich basis. The next result, due to Kalenda, provides nontrivial examples of Plichko spaces of continuous functions.

~~Theorem 17.14 (Kalenda [216]) Let K be a compact Abelian group. Then C(K) is 1-Plichko.~~

Proof We first show that the complex C(K) is 1-Plichko. After that, it is easy to deduce that the real C(K) is 1-Plichko, too (the converse implication is an open problem). Let G be the Pontryagin dual of K. Since the circle group is a subset of the complex field, G may be regarded as a subspace of C(K).Giveng ∈ G, define g∗(f ) = f · gdμ, K ∗ where μ is the Haar probability measure on K. Then g ∈ C(K) . Note that, given ∈ = = = = χ G, either K χdμ 1 when χ 1orelse K χdμ 0. Indeed, if χ 1, then 388 17 Banach Spaces with Many Projections choose a ∈ K such that χ(a)= 1 and observe that χ(x)dμ= χ(a + x)dμ = χ(a)χ(x)dμ= χ(a)· χ(x)dμ. K K K K = The first equality holds because μ is translation-invariant. Thus K χdμ 0. It follows that if f,g ∈ G and f = g, then g∗(f ) = 0 and g∗(g) = 1. Observe that, by the Stone–Weierstrass theorem, G is linearly dense in C(K) since it separates points, contains the constant one function and is closed under multiplication ∗ and complex conjugation. Thus, {(g, g )}g∈G is a Markushevich basis. It remains to show that it is countably 1-norming. Let Φ : L1(μ) → C(K) be the linear operator defined by Φ(h)f = f hdμ, f ∈ C(K) . K In particular, Φ(g) = g∗ for g ∈ G. Note that Φ is an isometric embedding. Since μ is strictly positive, im Φ is 1-norming for C(K) (note that C(K) is canonically embedded into L∞(μ), and L1(μ) is 1-norming for that space). Note that C(K) is formally a subspace of L1(μ) and is dense with respect to the L1 norm. Let H be the linear span of G. As we have already noticed, H is dense in C(K) with respect to the maximum norm. Thus, H is also dense in C(K) with respect to the L1 norm inherited from L1(μ). It follows that H is norm dense in L1(μ). Consequently, ΦH is norm dense in L1(μ) and therefore 1-norming for C(K). Finally, observe that ΦH is the linear span of {g∗ : g ∈ G} and the space

~~ {ϕ ∈ C(K) :|{g ∈ G: ϕ(g) = 0}| ≤ ℵ0} is norm closed in C(K). This completes the proof.~~

~~17.9 Preservation of Plichko spaces~~

Below is the announced preservation result involving the notion of a projectional sequence. We shall use it for characterizing Plichko spaces in terms of projectional skeletons. Finally, we show that every Plichko space has a countably norming Markushevich basis.

~~Theorem 17.15 (Kubis[´ 250]) Let {Pα}α<κ be a projectional sequence in a Banach space E, and let D ⊆ E be a norming space such that = D PαD α<κ~~

~~ and (PαE,PαD) is a Plichko pair for each α<κ. Then (E, D) is a Plichko pair. 17.9 Preservation of Plichko spaces 389~~

~~Note that we do not assume that the projections above are uniformly bounded. Before the proof, we need an auxiliary fact.~~

~~Lemma 17.7 Let (E, D) be a Plichko pair, and let P : E → E be a bounded projection such that P D ⊆ D. Then (ker P,D∩ ker P ) is a Plichko pair.~~

~~Proof Let A be a linearly dense subset of E such that~~

~~| supp(y, A)|≤ℵ0 for every y ∈ D.LetB ={a − Pa: a ∈ A}. Then B is linearly dense in~~

~~ker P = im(id E − P).~~

~~Let H = D ∩ ker P .Fixy ∈ D with y=1. Let z = y − P y. Then z ∈ D because P preserves D. Furthermore, z ∈ ker P and~~

~~z≤1 +P .~~

~~Given x ∈ ker P ,wehave~~

~~|z(x)|=|y(x) − (P y)x|=|y(x) − y(Px)|=|y(x)|.~~

~~This shows that H is norming for ker P . Finally, given y ∈ H ,wehave~~

~~y(a − Pa)= y(a) − y(Pa) = y(a) − (P y)a = y(a), so supp(y, B) = supp(y, A). Thus, B witnesses that (ker P,H) is a Plichko pair.~~

~~Now we prove Theorem 17.15.~~

~~Proof We construct inductively a family of sets {Aα}α<κ such that:~~

(i) Aα is a linearly dense subset of PαE; (ii) α<β⇒ Aα ⊆ Aβ ; (iii) supp(y, Aα) is countable for every y ∈ D; and (iv) Pαa = 0 whenever a ∈ Aβ \ Aα. We start with a linearly dense set A0 ⊆ P0E witnessing that (P0E,P D) is a Plichko 0 pair. We must check (iii). Observe that a = P0a and consequently y(a) = (P y)a ∗ 0 for every a ∈ A0 and y ∈ E . Hence = supp(y, A0) supp(P0y,A0) is countable and (iii) holds. Now, ﬁx an ordinal δ>0 and assume

~~{Aα}α<δ has already been deﬁned. 390 17 Banach Spaces with Many Projections~~

~~Suppose ﬁrst that δ = β + 1. We use Lemma 17.7.LetY = PδE, and let~~

~~D ={y Y : y ∈ D}⊆Y .~~

~~ Then (Y, D ) is a Plichko pair and Pβ Y is a projection whose dual preserves D . By Lemma 17.7, there is a linearly dense subset B of~~

~~ker Pβ ∩ Y witnessing that~~

~~(ker Pβ ∩ Y,E) is a Plichko pair, where E = D ∩ker[(Pβ Y)]. It follows that supp(y, B) is countable whenever ∈ ∩ y D ker Pβ .~~

Deﬁne Aδ = Aβ ∪ B. Conditions (i), (ii) and (iv) are obviously satisﬁed. It remains to check (iii). Fix y ∈ D. By the induction hypothesis, supp(y, Aβ ) is countable. Let = − z y Pβ y. Then

z ∈ D ∩ ker Pβ , so supp(z, B) is countable. Finally, given b ∈ B,wehave = − = − = z(b) y(b) (Pβ y)b y(b) y(Pβ b) y(b), and therefore supp(y, B) = supp(z, B) is countable. This shows (iii) because

~~supp(y, Aδ) = supp(y, Aβ ) ∪ supp(y, B).~~

~~Suppose now that δ is a limit ordinal. Deﬁne Aδ = Aξ . ξ<δ~~

Clearly, conditions (i) and (ii) are satisﬁed. Condition (iv) is obvious, so it remains to check (iii). Fix y ∈ D. There is nothing to prove if δ has a countable coﬁnality because then supp(y, Aδ) = supp(y, Aξ ). ξ<δ

Assume cf δ>ℵ0. Since Aδ ⊆ PδE, we see that = supp(y, Aδ) supp(Pδy,Aδ). = Thus, we may assume that y Pδy. Continuity of the projectional sequence implies that = y lim Pξ y, ξ<δ 17.9 Preservation of Plichko spaces 391

∗ where the limit is with respect to the weak topology. We know that Pδ is contained in a Σ-space, and therefore it is weak∗ countably tight. It follows that there exists = ≤ α<δsuch that Pξ y y for α ξ<δ. In particular,

~~supp(y, Aδ) = supp(y, Aα) because if a ∈ Aδ \ Aα, then~~

= = = y(a) (Pαy)a y(Pαa) 0, by (iv). By the induction hypothesis, supp(y, Aδ) is countable. This shows (iii). Finally, set A = Aα. α<κ By (i), A is linearly dense in E. It remains to check that supp(y, A) is countable ∈ ∈ ∈ for every y D.Fixy D. By the assumption, y PαD for some α<κ. Thus = Pξ y y whenever α ≤ ξ<κ, and we conclude as in the limit case of the construction above. This completes the proof.

~~We are now ready to characterize Plichko spaces in terms of projectional skeletons. A projectional skeleton {Ps}s∈Σ is commutative if~~

~~Ps ◦ Pt = Pt ◦ Ps holds for every s,t ∈ Σ.~~

~~Lemma 17.8 Let (E, D) be a Plichko pair, and assume D is a Σ-subspace of E. Then there exists a commutative projectional skeleton s ={Ps}s∈Γ on E such that D = D(s).~~

Proof Fix a big enough regular cardinal θ, and let Γ be the family of all countable elementary substructures M of (H (θ), ∈) such that (E, D) ∈ M. By Theorem 17.13 and the proof of Theorem 17.7,we know that s ={PM }M∈Γ is a projectional skeleton in E such that D ⊆ D(s), where ⊥ PM is induced by (E,D,M) (i.e., im PM = EM and ker PM = (D ∩ M)). Theo- rem 17.10(b) says that D(s) is weak∗ countably tight and that it is weak∗ countably closed (i.e., A ⊆ D whenever A ⊆ D is countable and the closure is in σ(E,E)). Thus D = D(s). 392 17 Banach Spaces with Many Projections

It remains to show that s is commutative. Given M ∈ Γ , deﬁne rM = PM G, where G ⊆ E is a linearly dense set witnessing that D is a Σ-space. We may assume that G ∈ M. We claim that x if x ∈ G ∩ M, r (x) = M 0ifx ∈ G \ M.

~~Indeed,~~

~~G ∩ M ⊆ (E ∩ M)= im PM . If x ∈ G \ M, then for y ∈ D ∩ M we have that~~

x/∈ supp(y, G) because supp(y, G) ⊆ M. Hence y(x) = 0fory ∈ D ∩ M, and therefore x ∈ ⊥ (D ∩ M) = ker PM . Using the property of rM above, we see that rM ◦ rN = rM∩N for M,N ∈ Γ . Since G is linearly dense in E, this shows that

~~PM ◦ PN = PM∩N = PN ◦ PM for every M,N ∈ Γ . This completes the proof.~~

Theorem 17.16 (Kubis[´ 250]) Let E be a Banach space, and let D ⊆ E be a k-norming subspace. The following are equivalent: (a) (E, D) is a Plichko pair. (b) E has a commutative projectional skeleton s such that D ⊆ D(s).

Proof (a) ⇒ (b) is contained in Lemma 17.8. (b) ⇒ (a): Suppose we have proved this implication for spaces of density <κ and ﬁx a Banach space E of density κ with a commutative k-projectional skeleton

~~{Ps}s∈Γ .~~

~~By (the proof of) Theorem 17.6, there exists a k-projectional resolution of the identity~~

p ={Pα}α<κ ⊆ = on E such that for each α<κthere is a directed set Sα Γ with Pαx lims∈Sα Psx for x ∈ E (to be formal, we need to assume that Γ ∩ κ =∅). Observe that, by continuity,

Ps ◦ Pα = Pα ◦ Ps holds for every s ∈ Γ and α<κ.LetD be the norming space induced by p.Fix ∈ ∈ = y D and ﬁx α<κ.Lets Γ be such that y Ps y. Then = = ∈ Pαy PαPs y Ps Pαy D. 17.9 Preservation of Plichko spaces 393

⊆ =ℵ Hence PαD D. Now use Theorem 17.15. In the case where cf κ 0,itmay happen that = D PαD, α<κ but we may replace D by PαD, α<κ still having a k-norming space. By Theorem 17.15, (E, D) is a Plichko pair.

Corollary 17.10 Let E be a Banach space of density ℵ1. The following properties are equivalent: (a) E is a 1-Plichko space. (b) E has a projectional resolution of the identity. (c) E = E , where {E } is a continuous chain of closed separable α<ω1 α α α<ω1 subspaces such that Eα is 1-complemented in Eα+1 for each α<ω1.

Proof Implication (b) =⇒ (c) is trivial and (c) =⇒ (b) is a consequence of Theo- rem 17.5. The fact that (a) =⇒ (b) follows from Theorem 17.6 together with Theo- rem 17.16. Finally, (b) =⇒ (c) follows again from Theorem 17.16 because a PRI of length ω1 is a commutative projectional skeleton.

~~We now turn to the announced result on Markushevich bases. This result is due to Kalenda (see [219, Theorem 4.15]), although the idea goes back to Valdivia [432].~~

~~Theorem 17.17 (Kalenda [219]) Given a Banach space E and k ≥ 1, the following conditions are equivalent: (a) E is a k-Plichko space. (b) E has a countably k-norming Markushevich basis.~~

Proof (b) =⇒ (a): Let {(xs,ys)}s∈Γ be a countably norming Markushevich basis, and let D consist of all y ∈ E such that {s : y(xs) = 0} is countable. By definition, D is norming and {xs : s ∈ Γ } witnesses that (E, D) is a Plichko pair. (a) =⇒ (b): We use induction on the density of E.IfE is separable, then we use Markushevich’s theorem (Theorem 17.12 above). Fix a k-Plichko space E of density κ>ℵ0, and assume that (a) =⇒ (b) holds for Banach spaces of density <κ. Actually, we need to have a stronger induction hypothesis: Given a k-Plichko pair (E, D), there is a countably k-norming Markushevich basis {(xs,ys)}s∈Γ such that ys ∈ D for every s ∈ Γ . Note that, by the proof of Markushevich’s theorem, this holds for any pair (E, D) where E is separable and D is norming. Let D ⊆ E be a Σ-subspace, and let {Pα}α<κ be a PRI induced by a commu- ⊆ = tative projectional skeleton in E. Then PαD D. Indeed, D s∈Σ Ps E , where {Ps}s∈Σ is a fixed commutative projectional skeleton. Given y ∈ D, find t ∈ Σ such = = = ∈ ⊆ that y Pt y. Then Pαy PαPt y Pt Pαy,soy Pt E D. 394 17 Banach Spaces with Many Projections

It sufﬁces to show that a countably k-norming Markushevich basis Mα = { } ∈ ∈ (xs,ys) s∈Γα in Eα such that ys D for s Γα can be extended to a Marku- M ={ } shevich basis α+1 (xs,ys) s∈Γα+1 in Eα+1 that is countably k-norming for = ∩ = ∩ Eα+1.LetF ker Pα Eα+1, and let H D ker(Pα Eα+1). By Lemma 17.7, (F, H ) is a Plichko pair. By an inductive hypothesis, there is a Markushevich basis {(xs,ys)}s∈T such that ys ∈ H for every s ∈ T . We may assume that T ∩ Γα =∅. = ∪ = ⊕ { } Set Γα+1 Γα T . Since Eα+1 Eα F , xs s∈Γα+1 is linearly dense in Eα+1. Formally, by the proof of Lemma 17.7, H is 2k-norming for Eα+1 but, on the + M other hand, PαE H is k-norming. Thus, ﬁnally, α+1 is countably k-norming for Eα+1.

~~We ﬁnish with the following result, perhaps ﬁrst noticed by Plichko (see Theo- rem 6.3 in [216] and the comment preceding it), which provides many examples of Plichko spaces.~~

~~Theorem 17.18 (Plichko) Every order-continuous Banach lattice is a 1-Plichko space.~~

~~Recall that a Banach lattice is order continuous if every decreasing net with inﬁ- mum zero converges to zero in the norm topology.~~

Proof We only sketch the idea of the proof. More details are given in the proof of Theorem 6.3 in [216]. Let E be an order-continuous Banach lattice. By [272, Proposition 1.a.9], E can be decomposed into an unconditional sum of mutually disjoint ideals, each having a weak unit. In fact, this is an 1-sum of such ideals. In order to conclude that E is 1-Plichko, it sufﬁces to show that each ideal in the decomposition is WCG. Theo- rem 17.15 implies that an arbitrary 1-sum of WCG Banach spaces is 1-Plichko. Let X be an order-continuous lattice with the weak unit e. This means that e ∧ x = 0 holds only if x = 0. The interval [0,e] is weakly compact by [272, Theo- ≥ = ∧ = x ∧ rem 1.b.16]. Furthermore, given x 0, let xn x ne. Then xn n( n e) belongs to the linear span of [0,e] and the sequence {xn}n∈ω is increasing, bounded by x. By order continuity, it converges to some y ∈ X. Since e is a weak unit, necessarily y = x. Thus, the closed linear span of [0,e] contains the positive cone of X, which shows that X is WCG and completes the proof.

~~Corollary 17.11 Given a compact K, the space C(K) is 1-Plichko.~~

~~Recall that C(K) spaces may not be Plichko. Perhaps the simplest counterexample is C(βN) = ∞.~~

~~Corollary 17.12 Given a semiﬁnite nonnegative measure μ, the space L1(μ) is 1-Plichko.~~

Recall that a measure μ is semiﬁnite if for every measurable set A with μ(A) = +∞ there is a measurable subset B ⊆ A such that 0 <μ(B)<+∞. Chapter 18 Spaces of Continuous Functions over Compact Lines

Abstract In this chapter, we discuss selected properties of Banach spaces of type C(K), where K is a linearly ordered compact space, called a compact line for short. In particular, we present Nakhmanson’s theorem stating that if K is a compact line such that Cp(K) is a Lindelöf space, then K is second-countable. We also discuss the separable complementation property in the context of compact lines. Compact lines are relatively easy to investigate, yet they form a rich class of spaces and provide several interesting examples. A very special case is the smallest uncountable well-ordered space ω1 + 1, which appeared several times in the previous chapters. Its space of continuous functions turns out to be a canonical example for several topological and geometric properties of Banach spaces. More complicated compact lines provide examples related to Plichko spaces.

~~18.1 General facts~~

In this chapter, we discuss selected properties of Banach spaces of type C(K), where K is a linearly ordered compact space, called a compact line for short. To be more precise, recall that a linearly ordered set X is a topological space when endowed with the interval topology generated by open intervals of the form (←,b) := {x ∈ X : x

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 395 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_18, © Springer Science+Business Media, LLC 2011 396 18 Spaces of Continuous Functions over Compact Lines

Proposition 18.1 Let f : K → L be an increasing surjection of linearly ordered spaces. If K is compact and f −1(p) is closed for every p ∈ L, then f is continuous. In particular, L is compact.

~~Proof Fix an increasing surjection f : K → L, where K and L are compact lines. Fix p ∈ L, and let q = min f −1(p) (here we use the compactness of K). Now~~

~~f −1[[p,→)]=[q,→).~~

~~Similarly, f −1[(←,p]] = (←,q]. Thus f is continuous because intervals of the form [p,→) and (←,p] (p ∈ L) form a closed subbase for the space L.~~

Recall that I denotes the real closed unit interval, a “generic” compact line in the sense that for every compact line K continuous increasing maps from K into I separate points. More precisely, we have the following lemma.

~~Lemma 18.1 Let K be a compact line a~~

Proof By Urysohn’s lemma, there is a continuous map g : [a,b] → I such that g(a) = 0 and g(b) = 1. Deﬁne f to be constant 0 on (←,a], constant 1 on [b,→), and f(t)= max g(s) s∈[a,t] for t ∈ [a,b]. It is clear that f is increasing, and it is easy to check that it is continuous.

~~Proposition 18.2 For every compact line K, continuous increasing functions on K with values in I form a linearly dense set in C(K).~~

Proof Let A consist of all such functions. Note that 1 ∈ A and, by Lemma 18.1, A separates the points of K. Furthermore, A is closed under multiplication (i.e., f · g ∈ A whenever f,g ∈ A). Let B be the linear span of A. Then B is a subring of C(K) that separates points and contains the constant functions. By the Stone– Weierstrass theorem, B is dense in C(K).

~~It is easy to determine when a compact line is connected: Its order must be dense in the sense that for every x~~

~~x ∼ y ⇐⇒ [ x,y] is scattered.~~

Since the property of being scattered is stable under subsets and ﬁnite unions, ∼ is clearly an equivalence relation with convex equivalence classes. It is not hard to check that the closure of an increasing sequence of scattered intervals is scattered, and therefore the equivalence classes of ∼ are closed intervals. Let L = K/ ∼ be endowed with the natural linear order (induced from K). By Proposition 18.1, the canonical increasing quotient map q : K → L is continuous and L is compact. Finally, L is connected because otherwise it would contain ajump[a,b], but on the other hand − − q 1(a) ∪ q 1(b) would be a scattered interval, which contradicts the fact that x ∼ y whenever q(x) = a and q(y) = b. Summarizing, we have the following proposition.

Proposition 18.3 Given a compact line K, there exist a connected compact line L and a continuous increasing quotient map q : K → L such that q−1(t) is scattered for every t ∈ L. Furthermore, for every connected compact line, for every continuous increasing map f : K → X there exists a unique increasing map h : L → X satisfying h ◦ q = f .

~~The map q : K → L will be called the connectiﬁcation of K. By the second statement, it is deﬁned uniquely up to an order isomorphism.~~

~~Proof Only the second statement requires an argument. Fix t ∈ L. Since q−1(t) is scattered, so is its image under f . Since X is connected, we conclude that~~

~~f [q−1(t)] is a singleton. Denote it by h(t). This shows that there is a unique way to deﬁne h so that h ◦ q = f . Clearly, h is increasing.~~

A similar procedure can be applied to a ﬁrst-countable compact line for getting a nowhere separable one. Recall that a topological space is nowhere separable if all its separable subsets are nowhere dense. For a linearly ordered space, this is equivalent to saying that nondegenerate intervals are not separable. 398 18 Spaces of Continuous Functions over Compact Lines

Proposition 18.4 Let K be a ﬁrst-countable compact line. Then there exists a continuous increasing quotient map q : K → L such that L is nowhere separable and q−1(t) is separable for every t. Furthermore, for every nowhere separable compact line, for every continuous increasing map f : K → X there exists a unique increasing map h : L → X satisfying h ◦ q = f .

Proof Deﬁne x ∼ y ⇐⇒ [ x,y] is separable. Then ∼ is an equivalence relation with convex classes. As K is ﬁrst-countable, the equivalence classes of ∼ are closed and separable. Let L = K/ ∼. Then L is a compact line and the natural quotient g : K → L is continuous and increasing. We need to show that L is nowhere separable. Suppose a

~~18.2 Nakhmanson’s theorem~~

~~The following result is due to Nakhmanson [309](seealso[27, Theorem IV.10.1]).~~

~~Theorem 18.1 Let K be a compact line such that Cp(K) is a Lindelöf space. Then K is second-countable.~~

Proof Let M consist of all continuous increasing maps of the form f : K → I. Clearly, M is a closed subset of C(K) with the pointwise topology. Fix f ∈ M.We are going to define a neighborhood U(f)∈ τp. 1 ∈ [ ] Suppose first that 2 f K , and let 1 F = f −1 . 2 Then F is a closed interval of the form [a,b] (possibly a = b). Define 1 3 U(f)= x ∈ M : x(a) > and x(b) < . 4 4 Suppose now that 1 ∈/ f [K]. 2 18.3 Separable complementation 399

~~= ← ∪ → 1 1 Then K ( ,a] [b, ), where a 2 . Deﬁne 1 U(f)= x ∈ M : x(a) <~~

~~In both cases, U(f) is a τp-neighborhood of f in M. Applying the Lindelöf property, there is a sequence~~

{fn}n∈ω ⊆ M = such that M n∈ω U(fn). We shall check that fn separates the points of K, which will prove that K is second-countable. Fix p 4 and g(b) < 4 . Since g is increasing, we conclude that p

~~1 1 and therefore fk(p) < 2 and fk(q) > 2 . 1 ∈ [ ] 1 ≤ ≤ If 2 / fk K , then g(a) < 2~~

~~1 fk(p) <~~

~~In both cases, fk(p) < fk(q), which completes the proof.~~

The proof above trivializes when K is zero-dimensional. Namely, in that case the set of all continuous increasing maps from K into {0, 1} is closed and discrete in Cp(K). On the other hand, the cardinality of this set equals the weight of K. Thus the Lindelöf number of Cp(K) equals the weight of K. The last fact holds for arbitrary compact lines, which was actually stated and proved by Nakhmanson [309]. In fact, our proof can be easily adapted to show the general situation. We have decided to present Nakhmanson’s theorem in the special case of Lindelöf number ℵ0 since we shall not use the general version.

~~18.3 Separable complementation~~

We have already seen a negative result concerning the existence of averaging operators for increasing quotients of compact lines (see Theorem 17.4). It turns out that, looking at the Lindelöf property of the weak topology, one can ﬁnd an example of a C(K) space, with K a compact line, that fails the separable complementation property. 400 18 Spaces of Continuous Functions over Compact Lines

~~We need to recall the notion of a double arrow line.FixasetA ⊆ (0, 1). Usually it is supposed to be dense and uncountable. Deﬁne D(A) to be the set~~

(I ×{0}) ∪ (A ×{1}) endowed with the lexicographic order. The space D(A) is easily seen to be compact. It is zero-dimensional whenever A is dense in (0, 1). It is homeomorphic (even order-isomorphic) to the Cantor set exactly in the case where A is countable and dense in (0, 1). Otherwise it is not second-countable and is called a double arrow line/space (induced by A). A double arrow space is sometimes called the split interval, especially where A = (0, 1). Given a double arrow K = D(A), the space C(K) has a copy of C(I) induced by the obvious quotient map q : K → I, where q(x,i) = x for (x, i) ∈ K. We shall say that q∗C(I) is the natural copy of C(I) in C(K). The following result origi- nates in Corson [103]. We actually repeat some of the arguments from the proof of Theorem 8.1 using slightly different language.

Theorem 18.2 Let A ⊆ (0, 1) be uncountable, and let K = D(A) be the double arrow induced by A. The natural copy of C(I) in the space C(K) is not contained in any separable complemented space. In particular, C(K) fails to have the separable complementation property.

~~Note that Theorem 17.4 already implies that the natural copy of C(I) is not complemented in C(K), even without assuming that A is uncountable.~~

Proof We first need to identify the space C(K) /C(I). Namely, we claim that it is − isomorphic to c0(A ), where A ={0}∪A.Givena ∈ A, define a = (a, 0) and a+ = (a, 1).Givenf ∈ C(K), define

~~(Tf )(a) = f(a+) − f(a−).~~

Finally, let (Tf )(0) = f(0). Observe that Tf ∈ c0(A ). Indeed, given ε>0, there is a ﬁnite cover U of K consisting of clopen intervals such that the oscillation of f on each member of U is <ε.Now,if

~~{a−,a+}⊆U~~

for some U ∈ U, then |(Tf )(a)| <ε. It follows that T : C(K) → c0(A ). It is clear that T is linear and T ≤2. Furthermore, f ∈ ker T if and only if f is constant on each set of the form {a−,a+}, where a ∈ A. This means that f = f ◦ q, where q : K → I is the canonical quotient (i.e., f belongs to the natural copy of C(I)). On the other hand, given v ∈ c0(A ),letf be the map on K deﬁned by f(t)= v(a). a∈A,a+≤t 18.3 Separable complementation 401

It is easy to check that f is continuous and ﬁnite valued and that Tf = v. Thus T is a surjection. Now suppose that P : C(K) → C(K) is a bounded projection such that im P is separable and ker T ⊆ im P . Then

C(K) = im P ⊕ Y, where Y = ker P . Then T Y is an isomorphic embedding. This gives a contradiction since Y , being isomorphic to a subspace of c0(A ), is weakly Lindelöf determined, and consequently C(K) would be weakly Lindelöf, contrary to Theo- rem 18.1.

In the original work of Corson [103], as well as in the proof of Theorem 8.1,the space of all bounded right-continuous functions that have a ﬁnite left limit at each point has been considered. This space can be naturally identiﬁed with the space of all continuous functions on the split interval. We now turn to a positive result on separable complementation. Recall that a linear operator T : C(K) → C(K) is regular if T 1 = 1 and Tf ≥ 0 whenever f ≥ 0. Notice that a regular operator necessarily has norm ≤ 1. We shall say that C(K) has the regular separable complementation property (regular SCP, for short) if for every separable subset A there exists a regular projection onto a separable subspace containing A.

~~Theorem 18.3 (Kubis[´ 252]) Let K be a compact line in which all separable subsets are second-countable. Then C(K) has the regular separable complementation property.~~

Before proving this theorem, we need to develop some properties of continuous functions on compact lines. Let K be a compact line, and let f : K → R be a function. We say that p ∈ K is irrelevant for f if one of the following conditions holds:

~~(1) p = 0K and f [p,b] is constant for some b>p. (2) p = 1K and f [a,p] is constant for some a~~

~~Lemma 18.2 Let f : K → I be a continuous increasing function, where K is a compact line. Then ess(f ) is separable.~~

~~Proof Let Y = ess(f ), and let~~

~~Z = f [Y ]=f [K]. 402 18 Spaces of Continuous Functions over Compact Lines~~

Observe that f Y is an increasing two-to-one map onto a second-countable linearly ordered space Z ⊆ R. Fix a countable dense set D ⊆ f [Y ] that contains all external points of f [Y ]. Then D = Y ∩ f −1[D] is countable. We claim that D is dense in Y . Fix a nonempty open interval U ⊆ Y . If f U is not constant, there are x

f −1(r) ⊆ D ∩ U, and therefore D ∩ U = ∅.If(f(x),f(y)) is empty, then f(y)∈ D; hence y ∈ D ∩ U. If f U is constant, then U ={x,y} with x ≤ y.Ify = 1Y , then y ∈ D ;if − x = 0Y , then x ∈ D . Suppose that 0Y

~~Let X be a closed subset of a compact space K.Anextension operator is a linear operator T : C(X) → C(K) satisfying (Tf ) X = f for every f ∈ C(X).~~

Lemma 18.3 Let X be a closed subset of a compact line K. Then there exists a regular extension operator T : C(X) → C(K) such that every f ∈ C(K) satisfying ess(f ) ⊆ X belongs to the range of T .

~~Proof For each a ∈ L (X), choose a continuous increasing function~~

~~− ha :[a ,a]→I~~

− such that ha(a ) = 0 and ha(a) = 1. Deﬁne ⎧ ⎪ ∈ ⎨⎪f(p) if p X, f(0 ) if p<0 , (Tf )(p) = X X ⎪f(1X) if p>1X, ⎩ − − (1 − ha(p))f (a ) + ha(p)f (a) if p ∈ a ,a for some a ∈ L (X) .

It is straight to check that Tf ∈ C(K) for every f ∈ C(X). Furthermore, (Tf ) X = f , T 1 = 1 and T is positive, and therefore it is a regular extension operator. Note that Tf is constant both on

~~[0K , 0X] and~~

~~[1X, 1K ] . 18.3 Separable complementation 403~~

~~Finally, if ess(f ) ⊆ X, then T(f X) is constant on each interval of the form [a−,a], where a ∈ L (X) \ L (K), and therefore f = T(f X). This shows that f is contained in the range of T .~~

Proof of Theorem 18.3 Fix a separable set A ⊆ C(K). By Proposition 18.2, there is a countable set F ⊆ C(K) consisting of increasing functions such that A is contained in the closed linear span of F .LetX be the closure of ess(f ). f ∈F

Notice that {g ∈ C(K) : ess(g) ⊆ X} is a closed linear subspace of C(K). Thus ess(g) ⊆ X for every g ∈ A.By Lemma 18.2, X is separable. By our assumption, X is second-countable and hence C(X) is a separable Banach space. Let

T : C(X) → C(K) be a regular extension operator satisfying the condition from Lemma 18.3. In particular, T(f X) = f for every f ∈ F . Finally, the composition of T with the restriction operator f → f X is a regular projection onto the range of T isomorphic to the separable space C(X).

One can prove more here: Under the assumptions of Theorem 18.3, the Banach space C(K) has the controlled separable projection property (CSPP). Indeed, it suffices to observe that the support of every regular Borel measure on a compact line is separable, and if T is a regular extension operator for X ⊆ K, then T fixes all measures whose support is contained in X. Recalling that ω1 + 1 satisfies the assumptions of Theorem 18.3, we obtain another proof of Corollary 12.9. An Aronszajn continuum is by definition a nonmetrizable first-countable compact line K in which all separable subsets are second-countable. Necessarily, K has weight ℵ1, and consequently C(K) has density ℵ1. Note that Aronszajn continua exist without extra set-theoretic assumptions. For more information on this subject, we refer to [407]. By a standard argument, passing to a quotient, we may assume that K is nowhere separable. More precisely, define a relation x ∼ y if and only if [x,y] is separable (in particular, second-countable). Then ∼ is an equivalence relation with closed convex classes. The quotient K/ ∼ is a nowhere separable Aronszajn continuum.

~~Theorem 18.4 Let K be a nowhere separable Aronszajn continuum. Then C(K) has the regular SCP, yet it is not a Plichko space. 404 18 Spaces of Continuous Functions over Compact Lines~~

Proof The ﬁrst part follows from Theorem 18.3. In order to show that C(K) is not Plichko, ﬁx a big enough regular cardinal θ and a countable M H(θ) such that K ∈ M. Then K/M is a second-countable compact line. Let q : K → K/M be the canonical quotient map. Clearly, the set

~~− {p ∈ K/M:|q 1(p)| > 1} is everywhere dense because otherwise there would exist a nontrivial interval [a,b] ⊆ K/M on which q−1 would be one-to-one; consequently~~

[q−1(a), q−1(b)] would be a nontrivial second-countable interval in K. By Theorem 17.4, q does not admit any averaging operator, and therefore (C(K) ∩ M) is not complemented in C(K). This shows that C(K) does not have a projectional skeleton and therefore is not Plichko. Chapter 19 Compact Spaces Generated by Retractions

Abstract This chapter presents several classes of nonmetrizable compact spaces that correspond to well-known classes of Banach spaces with many projections. In particular, we discuss the class of Valdivia compact spaces and its subclasses: Corson and Eberlein compact spaces. We discuss a general class of compact spaces obtained by limits of continuous retractive sequences. We also introduce the notion of a retractional skeleton, dual to projectional skeletons in Banach spaces. The last section contains an overview of Eberlein compact spaces with some classical results and examples relevant to the subjects of previous chapters.

~~19.1 Retractive inverse systems~~

Recall that a retraction in a topological space X is a continuous map r : X → X such that r ◦ r = r. Equivalently, r r[X]=idX. Related to a retraction is the notion of a right-invertible map: f : X → Y is right-invertible if there is a map j : Y → X such that r ◦ j = idY . In fact, if f ◦ j = idY , then j ◦ f is a retraction, and if f : X → X is a retraction, then the same map f treated as a surjection f : X → f [X] is right- invertible. Abusing notation, we shall sometimes say that f is a retraction if it is right-invertible and sometimes say that f is an internal retraction to stress that its range is a subset of its domain and its right inverse is inclusion. Inverse systems or sequences provide an important tool for investigating nonmetrizable compact spaces. We shall be mostly interested in retractive inverse systems, which are deﬁned below. Recall that an inverse system of topological spaces is a pair of the form

{ } { t } ( Xs s∈Σ , ps s,t∈Σ, s≤t ), where Σ is an up-directed partially ordered set, each Xs is a topological space and t ≤ each ps for s t is a map of Xt into Xs (called a bonding map), and the following conditions are satisﬁed: s = (1) ps idXs . r = t ◦ r ≤ ≤ (2) ps ps pt whenever s t r. ; t ; We shall denote this inverse system shortly by Xs ps Σ .

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 405 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_19, © Springer Science+Business Media, LLC 2011 406 19 Compact Spaces Generated by Retractions

In the language of category theory, an inverse system of topological spaces is just a contravariant functor from a directed poset category into the category of topological spaces. S = ; t ; Given an inverse system Xs ps Σ , its limit will be denoted, as usual, by S S ←−lim . Formally, lim←− consists of a topological space X and a family of maps

~~{ps : X → Xs}s∈Σ , called projections, satisfying the following condition:~~

(3) Given a space Y and a family {fs : Y → Xs}s∈Σ , there exists a unique map h: Y → X such that ps ◦ h = fs for every s ∈ Σ. The space X can be represented as the subspace of the product s∈Σ Xs consisting = t of all functions x satisfying x(s) ps(x(t)) for every s

S = ; t ; Lemma 19.1 Let Xs rs Σ be a retractive system of topological spaces, and = S : → ∈ ◦ = let X ←−lim . Then there are maps is Xs X (s Σ) such that rs is idXs for every s ∈ Σ, where rs : X → Xs is the projection.

∈ = S ={ ∈ Proof Fix u Σ. We shall use the fact that X ←−lim( Σ ), where Σ t : ≤ } { s } Σ u t . The family iu s∈Σ satisﬁes t ◦ t = t ◦ t ◦ s = s rs iu rs is iu iu 19.1 Retractive inverse systems 407 for every u

~~S = ; t ; Lemma 19.2 Let λ be an ordinal, and let Xs rs λ be a continuous inverse α+1 S sequence of topological spaces such that each rα is a retraction. Then is retractive.~~

Proof We use induction. Assume β<λand we have constructed appropriate maps η : → iξ Xξ Xη (ξ<η<β) satisfying conditions (4) and (5). = S If β is a limit ordinal, use Lemma 19.1 and the fact that Xβ ←−lim( β). Finally, β suppose that β = α + 1, and choose a right inverse k for rα . Deﬁne

~~β = ◦ α iξ k iξ . Then β ◦ β = α ◦ β ◦ ◦ α = rξ iξ rξ rα k iξ idXξ and, given ξ<η<β,wehave~~

~~β ◦ η = ◦ α ◦ η = ◦ α = β iη iξ k iη iξ k iξ iξ . This completes the proof.~~

~~It turns out that every totally disconnected compact space is the limit of a retractive system. This is a consequence of the following fact.~~

S = ; t ; Proposition 19.1 Assume Xs rs Σ is an inverse system of finite metric t S spaces such that each bonding map rs is a surjection. Then has a right inverse. = S : → Proof Fix a well ordering on X ←−lim . Each projection rs X Xs induces a partition of X into clopen sets. Let is(rs(x)) be defined as the -minimal element −1 ≤ of rs rs(x).Ifs t, then the partition induced by rt refines the one induced by rs . More precisely, −1 ⊆ −1 rt rt (x) rs rs(x) ≤ ∈ t = whenever s t and x X. Thus, setting is(p) rt (is(x)), we obtain a map t : → is Xt Xs t ≤ ≤ u = u t that is a right inverse of rs . Finally, if s t u, then is it is . S S Recall that spaces of the form←− lim , where is as above, form precisely the class of totally disconnected compact spaces. 408 19 Compact Spaces Generated by Retractions

Sometimes we shall need to represent a retractive inverse system “inside” a compact space as a certain collection of internal retractions. The way to do it is explained in the following two lemmas [255].

~~S = ; t ; Lemma 19.3 Let Xs rs Σ be an inverse system of topological spaces with { t } = S = ◦ a right inverse is s,t∈Σs~~

~~(1) s ≤ t ⇒ Rs ◦ Rt = Rs = Rt ◦ Rs . (2) x = lims∈Σ Rs(x) for every x ∈ X.~~

Proof Property (1) is clear. Fix a neighborhood u of x. We can assume that u = −1 (rs) [v] for some s ∈ Σ and an open set v ⊆ Xs .Fixt ≥ s. Then = ◦ = t ◦ ◦ ◦ = t ◦ = ∈ rs(Rt (x)) rs(it rt (x)) rs rt it rt (x) rs rt (x) rs(x) v.

~~Thus Rt (x) ∈ u for every t ≥ s. This shows (2).~~

Lemma 19.4 Assume {Rs : s ∈ Σ} is a family of internal retractions of a compact space K such that Σ is a directed partially ordered set and conditions (1) and (2) = ; t ; = [ ] t = of Lemma 19.3 hold. Then K ←−lim Ks Rs Σ , where Ks Rs K and Rs Rs Kt .

~~Proof If r~~

~~s ◦ t = ◦ = ◦ = = t Rr Rs (Rr Ks) (Rs Kt ) (Rr Rs) Kt Rr Kt Rr , S = ; t ; t [ ]= and thus Ks Rs Σ is indeed an inverse system, and if s~~

~~{Rs : s ∈ Σ} is compatible with S, which determines a unique continuous map~~

: → S f K ←−lim such that ◦ = Rs f Rs for every s ∈ Σ. Condition (2) implies that f is one-to-one. It remains to show that [ ]= S ∈ S f K ←−lim .Fixy ←−lim , and choose ∈ { : ≥ } x Rs(y) s t . t∈Σ 19.2 Monolithic sets 409

~~The set above is nonempty by the compactness of K and the directedness of Σ. Observe that if t~~

~~Later we shall need the following simple factorization property of σ -directed retractive inverse systems.~~

Lemma 19.5 Let {rs}s∈Γ be an inverse system of internal retractions of a compact space K (in the sense of Lemma 19.4) such that K is its limit and Γ is a σ -directed poset. Then, for every continuous map f : K → L with L second-countable, there exists t ∈ Γ such that

~~f = f ◦ rt .~~

~~Proof Fix a big enough cardinal θ and ﬁx a countable M H(θ)so that f ∈ M and {rs}s∈Γ ∈ M. By elementarity, M “knows” a ﬁxed, countable base of L.Lett ∈ Γ be such that s~~

rt (x) = rt (y) =⇒ f(x)= f(y) for every x,y ∈ K.Fixx,y so that f(x)= f(y). Find disjoint basic open sets U,V ⊆ L that are elements of M and such that f(x)∈ U, f(y)∈ V and U¯ ∩V¯ =∅. Observe that

~~(∃ s ∈ Γ)rs[U]∩rs[V ]=∅. (19.1)~~

Indeed, otherwise for each s ∈ Γ choose ps ∈ rs[U]∩rs[V ]; by compactness, ﬁnd an accumulation point p of the net {ps}s∈Γ . Knowing that p = lims∈Γ rs(p) we deduce that p ∈ U¯ ∩ V¯ , a contradiction. Now observe that all parameters of the formula (19.1)areinM. Thus, by elementarity, there is s ∈ Γ ∩ M satisfying (19.1). In particular, rs(x) = rs(y). Finally, rt (x) = rt (y) because s

~~19.2 Monolithic sets~~

A subset D of a topological space X is monolithic in X if for every countable A ⊆ D the space A (closure in X) is second-countable and is contained in D. This notion was introduced by Arkhangel’skii (see [27]) when studying dyadic compact 410 19 Compact Spaces Generated by Retractions spaces. In the literature, a “monolithic set” is usually called “ω-monolithic” or “ℵ0- monolithic.” It is obvious how to generalize this notion for larger cardinals in place of ℵ0. Also, often in the deﬁnition above it is asserted that A must have a countable network instead of a countable weight. Since we are going to use this notion for compact spaces only, there will be no confusion. Denote by M the class of all compact spaces that have a dense monolithic set.

~~Proposition 19.2 Class M is closed under arbitrary products, ﬁnite sums and continuous images.~~

~~Proof It is clear that M is closed under ﬁnite sums. Let~~

{Ks : s ∈ S} M = ∈ be a collection of spaces in , and let K s∈S Ks . For each s S, choose a dense monolithic set Ds ⊆ Ks . Also, ﬁx ds ∈ Ds . Deﬁne D = x ∈ Ds :|{s ∈ S : x(s) = ds}| ≤ ℵ0 . s∈S

Then D is dense in K, and it is easy to see that it is monolithic. Now assume K ∈ M and f : K → Y is a surjection. Then f [D] is monolithic whenever D ⊆ K is monolithic. Thus M is closed under continuous images.

~~Proposition 19.3 Let K ∈ M , and let E ⊆ K be a countable dense set such that t(p,K)≤ℵ0 for every p ∈ E. Then K is second-countable.~~

~~Proof Let D ⊆ K be dense and monolithic. Since each p ∈ E is in the closure of ≤ℵ0 many elements of D,wehaveE ⊆ D. Thus w(K) = w(E) ≤ℵ0.~~

~~Corollary 19.1 If βX ∈ M , then X is pseudocompact.~~

Proof It is well known that a nonpseudocompact space admits a continuous map onto a dense subset of R that extends to a continuous map from its CechÐStoneˇ compactiﬁcation onto βR. Thus it is enough to show that βR ∈/ M . This follows ℵ from Proposition 19.3 because w(βR) = 2 0 and Q is a dense countable subset of βR that consists of Gδ points.

~~Finally, we note the following corollary.~~

~~Corollary 19.2 Let K ∈ M be a compactiﬁcation of N. Then K is metrizable. 19.3 Classes R and RC 411~~

~~19.3 Classes R and RC~~

In this section, we discuss a general class of compact spaces obtained by limits of continuous retractive sequences. We also introduce the notion of a retractional skeleton, dual to projectional skeletons in Banach spaces. Class R, introduced in [73], is defined to be the smallest class of compact spaces that contains all metrizable ones and that is closed under limits of continuous retractive inverse sequences. There is a natural hierarchy on the class R. Namely, given a compact space K, define rkR(K) = 0 if and only if K is a metric compact. Furthermore, given a ≤ = S S = ; j ; positive ordinal β,letrkR(K) β if K ←−lim , where Ki ri λ is a continuous retractive sequence such that rkR(Ki)<β for every i<λ. Finally, define rkR(K) = β if and only if rkR(K) ≤ β and rkR(K) ≤ α for any α<β. Set rkR(K) =∞if and only if for no ordinal β we have that rkR(K) ≤ β. It is clear that

R ={K : rkR(K) < ∞}. Some properties of class R are valid for continuous images and even for a larger class defined in a similar spirit, adding continuous images. Namely, we define class RC as the smallest class that contains all metric compact spaces and is closed under continuous images and inverse limits of transfinite sequences of retractions. RC RC = RC ξ RC 0 Class can also be defined inductively as ξ∈Ord , where is the class of all metric compact spaces and RC β consists of all compact spaces = [ S] K f ←−lim , where S = ; β ; Kα rα κ is a continuous retractive sequence and each Kα belongs to RC ξ . ξ<β

~~β Given K ∈ RC, denote by rkRC (K) the minimal β such that K ∈ RC . We call rkRC (K) the RC-rank of K. We shall write rkRC (K) =∞if K/∈ RC. It is clear that~~

rkRC (K) ≤ rkR(K) for every compact K. Below is a simple statement on convergent sequences. The proof shows that the rank above together with transﬁnite induction is very useful. Later on, we shall prove a stronger statement concerning embedability of the ordinal segment [0,ω1] (i.e., the ordinal ω1 + 1).

~~Proposition 19.4 Every inﬁnite member of class RC contains a nontrivial convergent sequence. 412 19 Compact Spaces Generated by Retractions~~

Proof The statement is true if rkRC (K) = 0. Suppose β>0 and the statement is true for spaces of RC-rank <β.FixK ∈ RC with rkRC (K) = β. So, K = f [Y ], = S where Y ←−lim and S = ; η; Yξ rξ κ is a continuous inverse sequence of retractions such that rkRC (Yξ )<β for every ξ<κ.Ifsome

~~f [Yξ ] is inﬁnite, then by the induction hypothesis it contains a nontrivial convergent sequence, so assume that each f [Yξ ] is ﬁnite. Thus~~

{f [Yξ ]: ξ<κ} is an increasing sequence of ﬁnite subsets of K. This sequence does not stabilize, so cf(κ) =ℵ0. We may assume that κ =ℵ0. Now observe that K = f [Yn] n∈ω because otherwise K would be metrizable and we would have rkRC (K) = 0 <β. Fix x = f(y)∈ K such that x/∈ f [Yn] for any n ∈ ω. By the continuity of f and by Lemma 19.3(b), we have

~~x = lim f(rn(y)). n→∞~~

~~Also, f(rn(y)) = x for every n ∈ ω. Thus K contains a nontrivial convergent sequence.~~

~~19.4 Stability~~

It is easy to see that the classes R and RC are stable under typical functors. Below we prove a more precise statement involving rank. Recall that a functor F deﬁned on the category of compact spaces is continuous if it preserves limits of arbitrary inverse sequences. Typical examples are the probability measures functor P , the superextension functor λ and the Vietoris hyperspace functor exp. Another example is the functor that assigns to K the dual unit ball ∗ B(C(K)) with the weak topology.

~~Lemma 19.6 (Kubis[« 253]) Let F be a continuous covariant functor on compact spaces, and assume δ is an ordinal such that rkR(F (K)) ≤ δ for every compact metric space K. Then~~

~~rkR(F (K)) ≤ δ + rkR(K) for every compact space K. 19.4 Stability 413~~

Proof We use induction on α = rkR(K). The statement is true for α = 0. Assume = S S = ; η; K ←−lim , where Kξ rξ κ is a continuous retractive sequence such that rkR(Kξ )<αfor every ξ<κ. By the induction hypothesis,

~~rkR(F (Kξ )) ≤ δ + rkR(Kξ )<δ+ α.~~

~~Here we use the fact that δ + ξ<δ+ α whenever ξ<α(it may happen that~~

~~ξ + δ = α + δ for ξ<α). Notice that [S]= η F F(Kξ ), F rξ ,κ is again a continuous retractive sequence with limit F(K). Thus rkR(F (K)) ≤ δ + α.~~

~~Corollary 19.3 Let F be a continuous covariant functor on compact spaces such that F(X) ∈ R for every compact metric space X. Then F(K) ∈ R for every K ∈ R.~~

Proof There are continuum many homeomorphic types of metric compact spaces, and therefore there is an ordinal δ greater than all ordinals rkR(F (X)), where X is compact metric. Thus, Lemma 19.6 applies.

~~∈ R Corollary 19.4 Let K . Then the spaces P (K), λ(K), exp(K), BC(K) belong to R and have R-rank not greater than rkR(K).~~

~~Similar results hold for the class RC, as the following corollary shows.~~

Corollary 19.5 Let F be a continuous covariant functor on compact spaces that preserves surjections. Assume further that rkRC (F (K)) ≤ δ for every compact metric space K, where δ is a ﬁxed ordinal. Then

~~rkRC (F (K)) ≤ δ + rkRC (K) for every compact space K.~~

~~Proof Repeat the arguments of the proof of Lemma 19.6 using additionally the fact that F preserves surjections.~~

~~Corollary 19.6 Let F be a continuous covariant functor on compact spaces that preserves surjections and such that F(X)∈ RC for every compact metric X. Then F(K)∈ RC for every K ∈ RC.~~

~~∈ RC ∈{ } Corollary 19.7 Let K , and let L P (K), λ(K), exp(K), BC(K) . Then L ∈ RC and, moreover,rkRC (L) ≤ rkRC (K). 414 19 Compact Spaces Generated by Retractions~~

~~Corollary 19.8 Let X be a compact metric space. Then, for every compact space K, it holds that~~

~~rkR(K ⊕ X) ≤ rkR(K), rkR(K × X) ≤ rkR(K).~~

~~The same inequalities hold for rkRC as well.~~

~~Proof Let F(K)= K ⊕X and F(f)= f ⊕ idX (that is, F (f )(x) = f(x)for x ∈ K and F (f )(x) = x for x ∈ X). Furthermore, let~~

~~G(K) = K × X, G(f ) = f × idX~~

(that is, G(f )(p, x) = (f (p), x) for (p, x) ∈ K × X). It is clear that F and G are continuous covariant functors that preserve surjections and metric compact spaces. Thus, the inequalities above follow from Lemmas 19.6 and 19.5, respectively.

~~Theorem 19.1 (Kubis[« 253]) The classes R and RC are stable under arbitrary products and one-point compactiﬁcations of disjoint sums.~~

~~Proof An inﬁnite product is the limit of a continuous retractive sequence of smaller products, and therefore it sufﬁces to show that~~

~~K × L ∈ R~~

(K × L ∈ RC) whenever K,L ∈ R (respectively K,L ∈ RC). Fix L and define FL(K) = K × L. Then FL is a continuous covariant functor and by Corollary 19.8 we have that FL(K) ∈ R whenever K ∈ R and FL(K) ∈ RC whenever K ∈ RC. This shows the statement concerning products. Let K be the one-point compactification of Kα, α<κ and assume Kα ∈ R for each α. Define rα : K → K by rα(x) = x for x ∈ Kξ ξ<α and rα(x) =∞otherwise, where ∞ denotes the “point at infinity” that compactifies the disjoint sum. By Lemma 19.4, {rα}α<κ induces a continuous retractive sequence of one-point compactifications of disjoint sums ξ<α Kξ . Thus, by induction it is enough to show that K ⊕ L ∈ R whenever K,L ∈ R. However, this is proved by a similar argument as for products, using the functor GL(K) = K ⊕ L. Almost the same arguments can be applied to class RC.

~~The class R is obviously not stable under subspaces because all Tikhonov cubes are in R, while not all compact spaces are in R. However, we show the following. 19.5 Some examples 415~~

~~Theorem 19.2 (Kubis[« 253]) Let K ∈ R, and let U be a clopen subset of K. Then U ∈ R and rkR(U) ≤ rkR(K).~~

Proof Fix K ∈ R of rank β, and suppose the statement is true for compact spaces R ⊆ = S of -rank <β. Fix a clopen set U K, and let K ←−lim , where S = ; η; Kξ rξ κ is a continuous retractive sequence and rkR(Kξ )<βfor every ξ<κ. As V is clopen, there exist α<κ and a clopen set V ⊆ Kα such that U = −1 rα [V ]. Cutting off an initial part of S, we may assume that α = 0. Now let − = ξ 1[ ] η : → η Uξ (r0 ) V and, given ξ<η,letsξ Uη Uξ be the restriction of rξ . Then ; η; Uξ sξ κ is a continuous retractive inverse sequence with limit U. By the induction hypothesis, rkR(Uξ ) ≤ rkR(Kξ ) for every ξ<κ. Thus rkR(U) ≤ rkR(K).

~~19.5 Some examples~~

~~Below we present a few simple examples related to classes R and RC.Wealso prove a result on cutting P-points.~~

Lemma 19.7 Assume K ∈ R and p ∈ K. Let {pn}n∈ω be a one-to-one sequence disjoint from K, and let L = K ∪{pn}n∈ω be such that each pn is isolated and p = limn→∞ pn. Then L ∈ R and rkR(L) ≤ rkR(K) + 1.

~~Proof Let~~

~~Kn = K ⊕{p0,...,pn−1} and deﬁne rn : L → Kn by~~

~~rn Kn = idK and rn(pi) = p for i ≥ n. By Lemma 19.6,wehaverkR(Kn) = rkR(K). Observe that~~

~~rn ◦ rm = rm ◦ rn = rn for n~~

~~x = lim rn(x) n→∞ because for every x ∈ L there is n0 such that x = rn(x) for n ≥ n0. By Lemma 19.4, one gets that L is the limit of a retractive sequence induced by {rn}n∈ω. Thus~~

~~rkR(L) ≤ rkR(K) + 1. 416 19 Compact Spaces Generated by Retractions~~

~~By the lemma above, the linearly ordered space~~

−1 K = ω1 + 1 + ω , where X−1 denotes the line X with reversed order, belongs to class R. It turns out that the space + + −1 ω1 1 ω1 does not belong to class R. Below we prove a more general statement about it involving P-points. Recall that an element p of a topological space X is a P-point if it is not isolated and p ∈ int un n∈ω whenever {un}n∈ω is a sequence of neighborhoods of p. An element p ∈ X is a cutting P-point if X = A ∩ B so that {p}=A ∩ B and p is a P-point both in A and B. A typical example of a cutting P-point is

~~∈ + + −1 ω1 ω1 1 ω1 .~~

~~Theorem 19.3 (Kubis[« 253]) No space in class R can contain cutting P-points.~~

Proof Suppose otherwise, and let K ∈ R be a space of minimal R-rank that contains a cutting P-point p and let A,B witness this property. Using certain absolute- ness arguments from logic, we may assume that w(K) =ℵ1 (see [253, proof of Theorem 4.5]) for details. Let

~~{uα : α<ω1} denote a basis of p in K such that~~

~~α<β⇒ uβ ⊆ uα.~~

~~Such a sequence exists because p is a P-point in K and χ(p,K)=ℵ1. The assumptions on A,B mean in particular that both sets uα \ B and uα \ A are nonempty for every α<ω1. Now suppose~~

~~{rα : α<κ} is a sequence of internal retractions in K such that Kα = rα[K] has R-rank < rkR(K) for every α<κand K = Kα . α<κ 19.5 Some examples 417~~

~~Assuming κ is regular, we deduce that~~

~~κ ∈{ℵ0, ℵ1}.~~

~~Suppose κ =ℵ0. For each α<ω1, ﬁnd n(α) < ω such that~~

~~uα ∩ A ∩ Kn(α) =∅=uα ∩ B ∩ Kn(α).~~

~~This is possible because p is an accumulation point of both A and B and therefore uα \ B and uα \ A are nonempty open sets. Now there is n ∈ ω such that~~

{α<ω1 : n(α) = n} is uncountable. Then p is an accumulation point of both A ∩ Kn and B ∩ Kn, which implies that Kn is a counterexample to the theorem. This contradicts the minimality of rkR(K). It follows that κ =ℵ1. The same argument as above, using the minimality of rkR(K), shows that either

~~(∀ α<ω1)(∃ β<ω1)Kα ∩ uβ \ B =∅ (19.2) or the same holds for A in place of B. Moreover, if p/∈ Kα, α<ω1 then (19.2) holds for both A and B. Clearly, we may assume that either~~

(∀ α<ω1)rα(p) ∈ B (19.3) or the same holds for A in place of B. Thus, without loss of generality, we may assume that both (19.2) and (19.3) hold (interchanging the roles of A and B if necessary). Deﬁne uα = uξ+1. ξ<α Then u ⊆ u and u ={p}, and thus every neighborhood of p contains α α α<ω1 α some uα. Moreover, for a limit ordinal δ<ω1,wehave uδ = uξ = uξ . ξ<δ ξ<δ

~~Fix α<ω1. Observe that by (19.2) and (19.3)thesetB ∩ Kα is a neighborhood of rα(p) in Kα. By the continuity of rα, there is ξ(α)<ω1 such that~~

~~rα[uξ(α)]⊆B. (19.4) 418 19 Compact Spaces Generated by Retractions~~

~~On the other hand, the set uα \ B is open and nonempty, and therefore there is η(α) > α such that~~

~~Kη(α) ∩ uα \ B = ∅. (19.5)~~

~~Fix a limit ordinal δ<ω1 such that ξ(α) < δ and η(α) < δ whenever α<δ. Then, by (19.4), we have~~

~~rα[uδ]⊆B for every α<δand hence~~

~~rδ[uδ]⊆B because rδ(x) = limα<δ rα(x) for every x ∈ K. Now, for each α<δ,ﬁxqα ∈ Kδ ∩ uα \ B, which exists by (19.5). Let q be an accumulation point of the sequence~~

~~{qα : α<δ}.~~

~~Then q ∈ Kδ and q ∈ uα = uδ. α<δ~~

~~Furthermore, q = p because p is a P-point in A. Hence q/∈ B and rδ(q) = q, which contradicts the fact that rδ[uδ]⊆B.~~

~~19.6 The ﬁrst cohomology functor~~

It turns out that the first cohomology group of a space in R has a very simple structure, as we prove below. This will be used to show that Abelian compact groups in class R are products of metric compact spaces. In particular, this will show that class R is not stable under open maps. The results presented here come from [257] and [251]. The first cohomology functor H 1 = H 1( · , Z) in the sense of the theory of sheaves [184]ortheCechˇ theory [386, Chapter 6] has the following properties: (1) H 1(X) is countable whenever X is second-countable. (2) H 1(f ) is a monomorphism whenever f is a continuous surjection. (3) If = ; η; X ←−lim Xξ pξ κ , then H 1(X) is the inductive limit of the sequence of groups 1 1 η H (Xξ ), H pξ ,κ . 19.6 The first cohomology functor 419

For the proofs, we refer to [386]or[257]. Another important feature of the functor above is that H 1(G) coincides with the Pontryagin dual of G whenever G is a compact connected Abelian group (see [257] for more details).

~~Theorem 19.4 (Kubis[« 251]) Assume that K is a retract of a space from class R. Then H 1(K) is isomorphic to a direct sum of countable groups.~~

Before proving Theorem 19.4, we need two statements concerning discrete Abelian groups. Let us denote by H the class of all Abelian groups that are isomorphic to a direct sum of countable groups. We need the following two properties of this class.

{ } Lemma 19.8 Assume that Gα α<κ is a continuous increasing chain of subgroups = = = ⊕ of an Abelian group G such that G α<κ Gα, G0 0 and Gα+1 Gα Hα for every α<κ. Then G = Hα. α<κ

~~In particular, if {Hα : α<κ}⊆H , then also G ∈ H .~~

~~Proof Denote by G the algebraic sum Hα ⊆ G α<κ~~

~~(i.e., G consists of all elements of the form~~

~~x0 +···+xk−1, ∈ ⊆ ⊆ ⊆ where xi Hαi , i~~

~~0 <α0 <α1 < ···<αk and αi is a minimal ordinal α such that xi ∈ Hα (i ≤ k). By the continuity of the chain, we have αk = ξ + 1 and hence~~

~~x0 +···+xk−1 ∈ Gξ and xk ∈ Hξ . Thus xk = 0 because Gξ ∩ Hξ = 0. This shows that Hη ∩ Hα = 0 α=η for every η<κ. 420 19 Compact Spaces Generated by Retractions~~

~~Lemma 19.9 The class H is closed under direct summands. That is, if G = H ⊕ K ∈ H , then H ∈ H and K ∈ H .~~

Proof We use induction on the cardinality of the group. Of course, the claim is true for countable groups. Fix κ>ℵ0, and suppose that the statement holds for all groups of cardinality <κ.Fix G = Gα, α<κ where each Gα is a countable Abelian group. Let H be a direct summand of G, and let h: G → H be a group epimorphism such that h H = idH . Fix a regular cardinal χ big enough that h ∈ H(χ) and G ⊆ H(χ). Fix an increasing continuous chain

~~{Mα}α<κ of elementary substructures of (H (χ), ∈) such that h ∈ M0, H ⊆ Mα α<κ and |Mα| <κ for every α<κ.LetHα = H ∩ Mα. Then~~

{Hα}α<κ = is an increasing continuous chain of subgroups of H and H α<κ Hα. Fix α<κ, and let M = Mα. Observe that G ∩ M = Gα. ξ∈κ∩M ∈ ∩ ⊆ | |≤ℵ ⊆ Indeed, if ξ κ M, then Gξ M because Gξ 0. Thus ξ∈κ∩M Gξ M. ∈ ∩ = +···+ ∈ On the other hand, if x G M and x x0 xk−1, where xi Gξi , then

~~{ξi : i~~

~~Hα+1 = Hα ⊕ Kα, where Kα = Hα+1 ∩ ker(gα). Applying Lemma 19.8, we get H = H0 ⊕ Kα. α<κ 19.6 The ﬁrst cohomology functor 421~~

~~By the inductive hypothesis, H0,Kα ∈ H because both of these groups are direct summands of G ∩ Mα+1 and |G ∩ Mα+1| <κ. Hence H ∈ H , which completes the proof.~~

~~We are now ready to prove Theorem 19.4.~~

Proof By Lemma 19.9, it sufﬁces to show that H 1(K) ∈ H whenever K ∈ R. We use induction on rkR. By property (1) of the functor H 1, the statement is true for spaces of R-rank 0. Fix an ordinal β>0 and assume H 1(K) ∈ H whenever ∈ R = S = ; η; rkR(K) < β.FixK with rkR(K) β, and let Kξ rξ κ be a continuous = S retractive sequence with K ←−lim and such that rkR(Kξ )<βfor every ξ<κ.Let 1 1 G = H (K), and let Gξ = H (Kξ ). By properties (2) and (3), we may identify each Gξ with a subgroup of G so that

~~{Gξ }ξ<κ becomes a continuous increasing chain with G = Gξ . ξ<κ~~

~~Furthermore, each Gξ is a direct summand of G. By the inductive hypothesis, {Gξ : ξ<κ}⊆H . By Lemma 19.8,wehavethat G = G0 ⊕ Hξ , ξ<κ where Hξ is such that~~

~~Gξ+1 = Gξ ⊕ Hξ ,ξ<κ.~~

~~Since Hξ is a direct summand of Gξ , we deduce, using Lemma 19.9, that~~

~~{Hξ : ξ<κ}⊆H .~~

~~Thus G ∈ H .~~

Theorem 19.5 (Kubis[« 251]) Assume that G is a compact connected Abelian group that is at the same time a topological retract of some space from class R. Then G is isomorphic, in the category of topological groups, to a product of metrizable compact groups.

Proof Let H denote the Pontryagin dual of G. Since G is connected, H is discrete and torsion-free, and therefore isomorphic to H 1(G);see[257, Proposition 2.5] for a proof of this well-known fact. By Theorem 19.4, H can be decomposed into a direct sum of countable groups. Thus G is a product of metric groups because it is the dual of H and Pontryagin duality turns direct sums into products. 422 19 Compact Spaces Generated by Retractions

Corollary 19.9 (KubisÐUspenskij« [257]) There exists a compact connected Abelian group G of weight ℵ1 that is an open image of a product of metric compact spaces, while on the other hand G is not a retract of any space from class R.

Proof There exists a torsion-free Abelian group A of cardinality ℵ1 that is indecomposable [178, Sections 88 and 89]; that is, A has no proper direct summands. Let G be the Pontryagin dual of A. We claim that the compact group G has the required properties. The duals of torsion-free discrete groups are connected [205, Theorem 24.25], so X is connected. It is a well-known fact (for the proof, see [257, Proposition 2.5]) that the cohomology group H 1(G, Z) is isomorphic to A (in general, it is the torsion- free part of A). In particular, H 1(G, Z) is indecomposable. Theorem 19.4 implies that G is not a retract of any space from R. Finally, it is well known that every compact group is an epimorphic image of a product of a metric compact group and every epimorphism between compact groups is an open map.

One should compare the result above with Theorem 17.14: Denoting by G the compact group from Corollary 19.9, the Banach space C(G) is 1-Plichko, and consequently the space of probability measures P(G) as well as the dual unit ball of C(G) with the weak∗ topology are Valdivia compact.

~~19.7 Compact lines~~

We already know that the space + + −1 ω1 1 ω1 −1 is not in class R, while ω1 +1+ω ∈ R. Below, we characterize scattered linearly ordered spaces that are in R.LetK be a linearly ordered compact space. It is not hard to check that p ∈ K is a cutting P-point in K if and only if p is a P-point both in (←,p] and [p,→).

~~Theorem 19.6 ([253]) Let K be a scattered linearly ordered space. Then K ∈ R if and only if K has no cutting P-points.~~

Proof The necessity follows from Theorem 19.3. Fix a compact, scattered, linearly ordered space K with no cutting P-points. We use the idea from the proof of the Hausdorff theorem on the structure of scattered linearly ordered sets. Deﬁne the following relation on K:

~~x ∼ y ⇐⇒ (∀ a,b ∈[x,y]) [a,b]∈R.~~

~~We check that ∼ is an equivalence relation. Only transitivity requires an argument. So assume x~~

[a,b]=[a,y]⊕[y ,b]∈R by Theorem 19.1. Otherwise, reversing the order if necessary, we may assume that y = inf {yn}n∈ω, where {yn}n∈ω is strictly decreasing. Assume y0 = b. Since K is compact and zero-dimensional, we may also assume that each yn has an immediate successor zn−1 (i.e., yn

~~{bα : α<κ}⊆[a,b) = [ ]∈R be strictly increasing, continuous and such that b supα<κ bα. Then a,bα because a,bα ∈ C. Deﬁne rα :[a,b]→[a,bα] by~~

~~rα(x) = min{x,bα}.~~

Then {rα : α<κ} is an inverse sequence of internal retractions in [a,b]. Hence [a,b]∈R. This shows that x ∼ b for every x ∈ C (i.e., b ∈ C). By the same argument, inf C ∈ C (i.e., C is a closed interval). Finally, it remains to show that there is only one ∼-equivalence class. Suppose not, and ﬁx C,D ∈ K/ ∼ with max C = c

~~We have the following corollary.~~

Corollary 19.10 Let K be a scattered compact linearly ordered space. Then there exists a scattered, compact, linearly ordered space L ∈ R that has a two-to-one continuous order-preserving map onto K. In particular, K ∈ RC.

Proof Let L = K · 2, the lexicographic product of K by {0, 1}. Then L is scattered, compact and has no cutting P-points (every point of L is isolated from at least one 424 19 Compact Spaces Generated by Retractions side). Thus L ∈ R and of course the natural projection f : L → K is two-to-one and continuous.

~~It is not known how to characterize or classify compact linearly ordered spaces in classes R and RC. Below we prove two results concerning the structure of these spaces.~~

~~Theorem 19.7 Assume K ∈ R is a linearly ordered space and p ∈ K is a P-point in (←,p]. Then, for every q>p, the interval [p,q] is disconnected. In particular, ω1 +[0, 1] ∈/ R.~~

Proof Suppose the statement is not true, and let K and p ∈ K form a counterexample such that rkR(K) is minimal. Let q>pbe such that [p,q] is connected. Let λ denote the character of p in (←,p]. Then λ is an uncountable regular cardinal. Let

~~{rα : α<κ} be an inverse sequence of internal retractions in K such that~~

~~rkR(Kα)~~

= [ ] = for every α<κ, where Kα rα K . We may assume that κ cf κ and that there ∈ ∩ { } exists r (p, q) K0 (recall that α<κ Kα is dense in K). Let pα α<λ be a strictly = increasing continuous sequence with p supα<λ pα. Striving for a contradiction, we shall consider two cases. Case 1. κ = λ. For each α<λ, ﬁnd δ(α) < κ such that

~~Kδ(α) ∩ (pα,p)= ∅.~~

~~If κ<λ, then, by the regularity of λ,wehavethat~~

~~δ = sup δ(α) α<κ is an ordinal <λ.Ifκ>λ, then, by the regularity of κ, we can ﬁnd δ<λsuch that the set {α<κ: δ(α) = δ} has cardinality κ. In both cases,~~

~~p ∈ (Kδ ∩ (p0,p)), and therefore p is a P-point in~~

~~Kδ ∩ (←,p). 19.8 Valdivia and Corson compact spaces 425~~

On the other hand, A = rδ[[p,r]] is a closed and connected subset of Kδ that contains p,r. Thus [p,r]⊆A ⊆ Kδ. It follows that Kδ is also a counterexample to our statement, which contradicts the minimality of rkR(K). Case 2. κ = λ. Then κ>ℵ0. By Theorem 19.3, p ∈ Kα for some α<κbecause p must have a countable character in [p,q]. We may assume that p ∈ K0. As in Case 1, we deduce that [p,r]⊆K0. Thus, by the minimality of rkR(K), p is not a P-point in Kα for any α<κ. In other words,

~~p/∈ (Kα ∩ (←,p)) for every α<κ.Fixα<κ. By the continuity of rα at p = rα(p), there exists η(α) < κ such that~~

~~rα[(pα,p]] ⊆ [p,→).~~

~~Let ξ(α)>α be such that (pα,pα+1) ∩ Kξ(α) = ∅. Since~~

κ = cf κ>ℵ0, there exists δ<κsuch that η[δ]⊆δ and ξ[δ]⊆δ. Then pδ ∈ Kδ by the deﬁnition of the function ξ. Hence rδ(pδ) = pδ. On the other hand, by the deﬁnition of the function η,wehaverα(pδ) ≥ p for every α<δand hence also rδ(pδ) ≥ p. This is a contradiction.

~~19.8 Valdivia and Corson compact spaces~~

In this section, we present the class of Valdivia compact spaces, dual to Plichko spaces, and Corson compact spaces, dual to WLD spaces. We discuss their properties and stability and give some examples. A topological space is called Valdivia compact if it is homeomorphic to a subspace K of a cube [0, 1]κ so that K ∩ Σ(κ) is dense in K. A dense set D ⊆ K will be called a Σ-subset of K if there is an embedding j : K →[0, 1]κ such that D = j −1[Σ(κ)].

Proposition 19.5 Let K be an inﬁnite Valdivia compact space embedded in [0, 1]κ so that K ∩ Σ(κ) is dense in K. Then there is S ⊆ κ such that |S|=w(K) and supp(x) ⊆ S for every x ∈ K.

Proof Fix a sufﬁciently big cardinal χ and an elementary substructure M of H(χ) so that K ∈ M, |M|=w(K) and M contains a ﬁxed open basis of K.LetS = κ ∩M. Using Lemma 17.6 with f = idK , we conclude that x | S ∈ K and then x = x | S for every K. This just means that supp(x) ⊆ S for x ∈ K. 426 19 Compact Spaces Generated by Retractions

The proposition above implies that a Valdivia compact space of weight κ can be “properly” embedded into [0, 1]κ . We shall use this fact without any further reference. As we shall see, the class of Valdivia compact spaces is not stable under continuous (even open) images. It is not known whether it is stable under retractions (see Corollary 19.16 for a positive direction). Knowing that not all compact spaces are Valdivia, it becomes clear that this class is not stable under closed subspaces because all Tikhonov cubes are Valdivia. How- ever, we have the following positive result.

~~Proposition 19.6 The class of Valdivia compact spaces is stable under closed Gδ subspaces.~~

κ Proof Let K ⊆[0, 1] be such that Σ(κ)∩ K is dense in K, and ﬁx a closed Gδ- subset L of K. It sufﬁces to show that Σ(κ)∩ L is dense in L. { } [ ]κ ⊆ Fix a sequence Un n∈ω of open subsets of 0, 1 so that cl (Un+1) Un and ∩ = ⊆[ ]κ ∩ = ∅ K n∈ω Un L. Fix an open set V 0, 1 such that L V . For each n ∈ ω, choose xn ∈ Σ(κ) ∩ K ∩ Un ∩ V .Letx be an accumulation point of the { } ∈ ∈ ∩ = sequence xn n∈ω. Then x Σ(κ) and, moreover, x K n∈ω Un L. Finally, x is in the closure of V . By regularity, this shows that Σ(κ)∩ L is dense in L.

~~In order to get simple examples of compact spaces that are not Valdivia, it is sufﬁcient to observe the following property.~~

~~Proposition 19.7 Every Valdivia compact has a dense monolithic set.~~

~~Proof Every relatively closed subset of a Σ-product is monolithic.~~

By Corollary 19.2, nonmetrizable compactiﬁcations of the integers cannot be Valdivia compact. Recall that K is Corson compact if it is homeomorphic to a compact subset of a Σ-product Σ(κ) for some κ.

~~Proposition 19.8 A Valdivia compact space is Corson if and only if it has countable tightness.~~

Proof Every Σ-product of real lines has countable tightness by Proposition 17.9, and therefore Corson compact spaces are countably tight as well. Now assume K is a countably tight Corson compact and that K ⊆[0, 1]κ , where D = Σ(κ)∩ K is dense in K.Fixp ∈ K. Then p ∈ A for some countable A ⊆ D.Let S = supp(x). x∈A Then S is countable and supp(p) ⊆ S. Thus p ∈ D. It follows that K ⊆ Σ(κ),soit is Corson. 19.8 Valdivia and Corson compact spaces 427

The following properties are useful for showing that certain natural compact spaces are not Valdivia. Recall that a topological space X is FréchetÐUrysohn if for every p ∈ A ⊆ X there is a sequence {an}n∈ω ⊆ A such that p = limn→∞ an.

Theorem 19.8 Let K be a Valdivia compact space, and let D be its Σ-subset. Then D is countably tight and FréchetÐUrysohn, and the closure in K of every countable subset of D is metrizable and contained in D (i.e., D is monolithic in K). Furthermore, K is the CechÐStoneˇ compactiﬁcation of D.

Proof The fact that D is countably tight follows from Proposition 17.9. κ Let K ⊆[0, 1] be embedded so that D = Σ(κ) ∩ K, and ﬁx a countable set ⊆ = A D.LetS a∈A supp(a). Then S is countable and

K0 ={x ∈ K : supp(x) ⊆ S} is closed in K, contained in D and metrizable (homeomorphic to a subset of the Cantor space [0, 1]S ). Thus the same applies to A. Now let p ∈ B ⊆ D. Using the countable tightness, ﬁnd a countable A ⊆ B such that p ∈ A. By the previous argument, A is metrizable, and therefore p is the limit of a sequence from A. Finally, in order to show that K = βD, we use a well-known criterion: Disjoint relatively closed subsets of D must have disjoint closures in K. However, this follows immediately from Corollary 17.9.

~~The next result, relating Valdivia compact spaces to class R, was already proved by Argyros, Mercourakis and Negrepontis in [15] before the class of Valdivia was formally deﬁned and studied.~~

Lemma 19.10 Let K be a Valdivia compact space suitably embedded in [0, 1]κ . Then, for every inﬁnite set T ⊆ κ there exists S ⊆ κ such that T ⊆ S, |T |=|S| and x | S ∈ K for every x ∈ K, where x | S = x · χS .

~~Proof Fix an elementary substructure M of a suitably big H(χ) such that K ∈ M, T ⊆ M and |T |=|M|.LetS = M ∩κ. By Lemma 17.6, we have that x | S ∈ K = K for every x ∈ K.~~

~~Let K and S be as above, and let rS(x) = x | S. Then rS : K → K is an internal retraction and its range is~~

~~K | S ={x ∈ K : supp(x) ⊆ S}.~~

~~Observe that the space K | S is again Valdivia compact and~~

~~rS[Σ(κ)∩ K] is a Σ-subset of K | S. A standard transﬁnite induction gives the following theorem. 428 19 Compact Spaces Generated by Retractions~~

ℵ = S Theorem 19.9 Let K be a Valdivia compact of weight κ> 0. Then K ←−lim , S = ; η; where Kξ rξ κ is a continuous retractive sequence of Valdivia compact spaces such that w(Kξ ) ≤|ξ|+ℵ0 for every ξ<κ.

Proof We may assume that K ⊆[0, 1]κ so that Σ(κ)∩ K is dense in K (see Propo- sition 19.5). Construct inductively a chain of sets {Sα}α<κ such that α ⊆ Sα, |Sα|≤α, Sα = κ α<κ and each Sα satisﬁes the assertion of Lemma 19.10 (i.e., x | Sα ∈ K for every x ∈ K, α ∈ κ). Given a limit ordinal δ,wehave x | Sξ = lim x | Sξ , ξ<δ ξ<δ and therefore we may assume that Sδ = Sξ ξ<δ

~~(i.e., that the chain {Sξ }ξ<κ is continuous). Deﬁne rα(x) = x | Sα. Then~~

~~x = lim rξ (x), ξ<κ and for a limit ordinal δ we have~~

~~rδ(x) = lim rξ (x). ξ<δ~~

By Lemma 19.4, the sequence {rα}α<κ induces a continuous retractive sequence S = ; η; = S = | Kξ rξ κ with K ←−lim . Finally, the space Kξ K Sξ is Valdivia compact of weight less than

~~|Sξ |≤|ξ|+ℵ0. This completes the proof.~~

~~Corollary 19.11 Valdivia compact spaces belong to class R.~~

~~It is not known whether the class of Valdivia compact spaces is stable under retractions. We show below that retracts of Valdivia compact spaces belong to R.~~

Lemma 19.11 Let r : K → L be a retraction, where K is a Valdivia compact space. Let θ be a big enough regular cardinal, and let M (H (θ), ∈) be such that r ∈ M. Then the induced map rM : K/M → L/M is a retraction. 19.8 Valdivia and Corson compact spaces 429

Proof As we have already mentioned, f → f M is a functor on the category of continuous maps in M.Now,ifθ is big enough, then, by elementarity, M |= “r is right-invertible.” This means that there is j ∈ M such that r ◦ j = idL. Finally, M M M M r ◦ j = (r ◦ j) = (idL) = idL/M .

~~Corollary 19.12 Retracts of Valdivia compact spaces belong to class R.~~

Proof We use induction on weight. Fix a retraction r : K → L, and assume K is Valdivia compact, w(L) = κ and the statement is true for spaces of weight <κ.Fix a big enough regular cardinal θ so that r ∈ H(θ) and C(L) ⊆ H(θ).Letλ = cf κ, { and ﬁx a continuous chain Mα}α<λ of elementary substructures of (H (θ), ∈) so | | = that Mα <κ for each α<λand N α<λ Mα contains a point-separating subset of C(L). In particular, L/N = L and {Mα}α<λ induces a continuous in- Mα : → verse sequence with limit L and whose αth projection is qL L L/Mα.By Mα : → Mα ◦ = Mα ◦ Mα Lemma 17.6, qK K K/Mα is a retraction and we have qL r r qK . M Mα By Lemma 19.11, r is a retraction. Thus qL is a retraction, too. This shows that the inverse sequence induced by {Mα}α<λ is retractive. Finally, L/Mα has weight ≤|Mα| <κ and therefore L/Mα ∈ R by the induction hypothesis. Thus L ∈ R.

~~We shall show later that retracts of Valdivia compact spaces of weight ℵ1 are Valdivia compact (see Section 19.10).~~

~~Example 19.1 (a) Let~~

K1 = ω1 + 1 be considered with the linear order. The space K1 is compact and can be represented ω as a subset of {0, 1} 1 consisting of all nonincreasing functions: An element α ∈ K1 corresponds to the characteristic function of [0,α). This shows that K1 is Valdivia. On the other hand, ω1 ∈ K1 witnesses that it is not countably tight and therefore not Corson. (b) Let −1 K2 = ω1 + 1 + ω , −1 where X denotes X with the reversed ordering. Clearly, K2 is a two-to-one image of K1. We claim that it is not Valdivia compact. For the contrary, suppose that D is a Σ-subset of K2. Then D is dense, so it contains all isolated points of K2.Onthe other hand, D is sequentially closed, so ω1 ∈ D. This is a contradiction because Σ- products are countably tight. On the other hand, notice that K2 ∈ R. More precisely, rkR(K2) = 2 (the fact that rkR(K2) ≤ 2 follows from Lemma 19.7, and we shall see later that spaces of R-rank 1 are Valdivia compact). Note that the space K2 is a two-to-one image of K1 = ω1 + 1. This example, showing that Valdivia compacta are not stable under continuous images, was ﬁrst noticed by Valdivia [433]. (c) Let = + + −1 K3 ω1 1 ω1 . 430 19 Compact Spaces Generated by Retractions

Then K3 is not Valdivia compact by Theorem 19.3. A direct argument is as follows: The only possible Σ-set D ⊆ K3 would have to consist of all countable ordinals and all reversed countable ordinals; on the other hand, βD is homeomorphic to + + −1 ω1 2 ω1 , not to K3. (d) Let

K4 = ω2 + 1, again considered as a compact linearly ordered space. We claim that it is not Valdivia κ −1 compact. Suppose f : K4 →[0, 1] is an embedding such that f [Σ(κ)] is dense. Note that f(α)∈ Σ(κ) whenever α has a countable coﬁnality. Thus,

~~f(ω2)/∈ Σ(κ) because of tightness. Since~~

f(ω2) = lim f(ξ) ξ<ω2 κ and the convergence in [0, 1] is pointwise, there is α0 <ω2 such that supp(f (ξ)) is uncountable for every ξ>α0. This is a contradiction because supp(f (α0 + 1)) is supposed to be countable. Notice that K4 ∈ R, being the limit of a continuous retractive sequence of Valdivia compact spaces. In particular, rkR(K4) = 2.

~~Valdivia compact spaces are dual to Plichko spaces. More precisely, we have the following theorem.~~

~~Theorem 19.10 (Kalenda [219]) For every Valdivia compact K, the Banach space C(K) is 1-Plichko. For every 1-Plichko space E, the dual unit ball BE with the weak∗ topology is Valdivia.~~

Proof Assume K ⊆[0, 1]κ so that D = K ∩ Σ(κ) is dense in K. For each α ∈ κ,letxα be the restriction to κ K of the projection of [0, 1] onto the αth coordinate. That is, xα(p) = p(α) for p ∈ K.Let ={ : ∈ } A xα1 xα2 ...xαn α1,α2,...,αn κ . By the StoneÐWeierstrass theorem, A is a linearly dense subset of C(K). Identifying K with a suitable subset of C(K) , it is clear that D is 1-norming and (A, D) is a Plichko pair. Now let E be a 1-Plichko space, and let

~~T : E → Rκ be a weak∗ continuous injective linear operator such that~~

~~D = T −1[Σ(κ)] 19.8 Valdivia and Corson compact spaces 431~~

∗ is 1-norming for E.LetK = BE with the weak topology. Then T K is a home- κ omorphic embedding of K into R . It remains to check that D0 = D ∩ K is dense. Suppose otherwise, and ﬁx y ∈ K \ (D0). Note that D0 is convex because D is a linear space. Thus, by the separation theorem, there is x ∈ E such that

~~y(x) > sup z(x). z∈K~~

~~But this contradicts the fact that D is 1-norming:~~

~~x = sup |d(x)|. d∈D∩K~~

~~For spaces of continuous functions, we have one more result.~~

~~Corollary 19.13 Let K be a compact space such that C(K) is a 1-Plichko space. Then the space of probability measures P(K)is Valdivia compact.~~

~~Proof By Theorem 19.10, BC(K) is Valdivia compact. By Proposition 19.6,itsuf- ﬁces to observe that P(K)is a (closed) Gδ-subset of BC(K) .~~

As we shall see later, P(K) is Valdivia whenever K is, but there are examples of non-Valdivia compact spaces for which P(K) is Valdivia (see Corollary 19.9 below). For certain applications, the following criterion for Valdivia compactness is useful. Recall that a family of sets A is T0-separating in X if for every x,y ∈ X there is A ∈ A satisfying |{x,y}∩A|=1. A family A is point-countable on D if the set

~~{A ∈ A : p ∈ A} is countable for every p ∈ D.~~

Theorem 19.11 Let K be a compact space. Then K is Valdivia if and only if there are a dense set D ⊆ K and a family U of open Fσ subsets of K that are T0- separating and point-countable on D. If K is totally disconnected, then the family U may consist of clopen sets.

~~Proof Assume K ⊆[0, 1]κ and D = Σ(κ)∩ K is dense in K. Deﬁne~~

r ={ ∈ : } Uα x K x(α) > r , U r ∈ and let consist of all sets of the form Uα, where α κ and r is a positive rational. Clearly, U consists of open Fσ sets and is T0-separating: Given x = y and ∈ ∈ r ∈ r assuming x(α) > y(α) for some α κ, we have that x Uα and y/Uα, where r ∈ ∩ r is a rational such that x(α)>r>y(α).Givenx D Uα, it must be the case that 432 19 Compact Spaces Generated by Retractions

α ∈ supp(x), so there are only countably many possibilities for α and r. Thus U is point-countable on D. Assume now that U is a T0-separating family of open Fσ subsets of K that is point-countable on a dense set D. For each U ∈ U , choose a continuous map −1 fU : K →[0, 1] such that U = f [(0, 1]]. The diagonal map U f : K →[0, 1] , deﬁned by

f (x)(U) = fU (x), is an embedding (because U is T0-separating) and f [D]⊆Σ(U ). This shows that K is Valdivia. Finally, suppose K is a totally disconnected Valdivia compact, and let U be a family of open Fσ sets having the properties above. Replace each U ∈ U by a countable sequence of clopen sets whose union is U. In this way, we obtain a family of clopen sets that is T0-separating and point-countable on the same set as U was.

~~A variant of the result above concerning Corson compact spaces reads as follows. The proof is exactly the same as for Theorem 19.11.~~

Theorem 19.12 Let K be a compact space. Then K is Corson if and only if there is a point-countable T0-separating family of open Fσ -subsets of K. Where K is totally disconnected, the family may be assumed to consist of clopen sets.

~~The following fact on zero-dimensional Valdivia compact spaces will be needed later.~~

Proposition 19.9 Let K be a Valdivia compact with a dense Σ-subset D. Then there exist a totally disconnected Valdivia compact, a dense Σ-subset E of L and a continuous surjection f : L → K satisfying f [E]=D.

Proof Let K ⊆ Iκ so that D = Σ(κ) ∩ K.Letφ : 2ω → I be a continuous surjection satisfying φ−1(0) ={0}. This map induces a surjection Φ : 2ω·κ → Iκ satisfying Φ−1[Σ(κ)]=Σ(ω · κ).LetE = Φ−1[D], and let L be the closure of E in 2ω·κ . Then L is a zero-dimensional Valdivia compact, E is its dense Σ-subset and Φ[E]=D, Φ[L]=K.

~~19.9 Preservation theorem~~

In order to get a better understanding of the class of Valdivia compact spaces, one needs to take care of their Σ-subsets. Namely, we shall consider pairs of the form (K, D), where D ⊆ K is dense and such that

j[D]⊆Σ(κ) 19.9 Preservation theorem 433 for some embedding j : K →[0, 1]κ . We shall call (K, D) a Valdivia pair.This is of course analogous to the notion of a Plichko pair. Using the corresponding preservation result for Plichko spaces, we obtain the following.

Theorem 19.13 (Kubis[« 250]) Let {rα}α<κ be a continuous retractive sequence in a compact space K. Let D ⊆ K be a dense set such that for each α<κ [ ] [ ] = [ ] (rα K ,rα D ) is a Valdivia pair and D α<κ rα D . Then (K, D) is a Valdivia pair.

Proof The sequence {rα}α<κ induces a projectional sequence {Pα}α<κ on the Ba- nach space C(K). In fact, identifying K with a suitable subset of the dual space [ ] C(K) , rα is just the restriction of Pα to K.Now(C(Kα) ,rα D ) is a Plichko pair, and the assumptions of Theorem 17.15 are satisﬁed. Using this theorem, we conclude that (C(K) ,D) is a Plichko pair. In particular, there is a one-to-one weak∗ continuous linear transformation T : C(K) → RΓ such that D ⊆ T −1[Σ(Γ)].This shows that (K, D) is a Valdivia pair.

~~Corollary 19.14 The limit of a continuous retractive sequence of Corson compact spaces is Valdivia compact.~~

~~Corollary 19.15 (KubisÐMichalewski« [255]) The limit of a continuous retractive sequence of metric compact spaces is Valdivia compact.~~

~~Corollary 19.16 (KubisÐMichalewski« [255]) Let K be a retract of a Valdivia compact. If w(K) ≤ℵ1, then K is Valdivia compact.~~

~~Proof Repeat the arguments from the proof of Corollary 19.12. We conclude that K is the limit of a continuous retractive sequence of metric compact spaces. Thus, K is Valdivia by Corollary 19.15.~~

~~Corollary 19.17 (KubisÐMichalewski« [255]) Let K be an open image of a Valdivia compact. If K is totally disconnected and w(K) ≤ℵ1, then K is Valdivia compact.~~

Proof As in the proof of Corollary 19.12, we obtain that K is the limit of an in- { } verse sequence of spaces K/Mα, α<ω1, where Mα α<ω1 is a continuous chain of countable elementary substructures of a big enough H(χ). We assume that M0 “knows” an open surjection r : L → K from a Valdivia compact L onto K. M Fix α<ω1, and let M = Mα. We claim that r is an open map. M [ −1[ ]] ∈ ∩ A basic open set in L/M is of the form qL f U , where f C(L) M and U = (0, →) ⊆ R. Thus f = f M and we have that

M [ [ −1 ]] = M [ M [ −1[ ]]] qK r f U r qL f U M = M because f f . The set on the right-hand side is open because qK is a quotient map and r[f [U]] is assumed to be open. 434 19 Compact Spaces Generated by Retractions

Thus rM is indeed an open map. Since K/M is totally disconnected and metrizable, a version of Michael’s selection theorem says that (rM )−1 has a continuous selection (i.e., rM is right-invertible). It follows, by the same arguments as in the M proof of Corollary 19.12, that qK is a retraction. Finally, K is Valdivia compact as the limit of a continuous retractive sequence of metric compact spaces.

It has also been proved in [255] that retracts of Tichonov or Cantor cubes are Valdivia compact. Recall again that it is not known whether the class of Valdivia compact spaces is stable under retractions. It is not stable under open continuous images because of Corollary 19.9.

~~19.10 Retractional skeletons~~

Below we introduce the notion of a retractional skeleton, dual to a projectional skeleton. We characterize Valdivia compact spaces in terms of retractional skeletons. We also present Bandlow’s characterization of Corson compact spaces in terms of elementary submodels. Fix a compact space K.Aretractional skeleton in K is a family of retractions {rs}s∈Γ of K satisfying the following conditions: (1) Γ is a σ -directed partially ordered set. (2) rs ◦ rt = rt ◦ rs = rs whenever s ≤ t. (3) Each rs[K] is metrizable. (4) x = lims∈Γ rs(x) for every x ∈ K. = = (5) Given s0

~~Theorem 19.14 (KubisÐMichalewski« [255]) A compact space is Valdivia if and only if it admits a commutative retractional skeleton.~~

Proof Suppose ﬁrst that K ⊆ Iκ is such that Σ(κ) ∩ K is dense in K.Callaset T ⊆ κ admissible if x · χT ∈ K for every x ∈ K. By Lemma 19.10, every countable set is contained in a countable admissible set. Denote by Γ the collection of all countable admissible sets, and for T ∈ Γ let rT (x) = x · χT . It is easy to see that {rT }T ∈Γ is a commutative retractional skeleton in K. Now assume that K has a commutative retractional skeleton {rs}s∈Γ , and let p = {Ps}s∈Γ be the dual projectional skeleton. By Theorem 17.16, C(K) is 1-Plichko = and (by the proof of Theorem 17.16) D s∈Γ im Ps is a Σ-subspace of the dual; that is, there is a one-to-one weak∗ continuous linear transformation T : C(K) → 19.10 Retractional skeletons 435 Rκ = −1[ ] [ ]⊆ such that D T Σ(κ) . In particular, s∈S rs K D is a dense Σ-subspace of K and T witnesses that K is Valdivia.

~~The “if” part of the statement above can also be proved directly by induction on the weight of K using Theorem 19.13.~~

~~Corollary 19.18 Let F be a continuous weight-preserving functor on compact spaces. Then F(K)is Valdivia whenever K is.~~

Proof A commutative retractional skeleton can be described as a pair (S, I), where S = ; t ; Ks rs Σ is a continuous retractive inverse system of metric compact spaces I = t S t ◦ t = and is,Σ is a direct system of right inverses to ; that is, rs is idKs and u ◦ t = u it is is holds whenever s

(is ◦ rs) ◦ (it ◦ rt ) = (it ◦ rt ) ◦ (is ◦ rs) holds for every s,t ∈ Σ, where is is the uniquely deﬁned embedding of Ks into K = S ◦ = t ←−lim satisfying rt is is for t>s. Now it is obvious that every continuous weight- preserving functor maps such a pair (S, I) to a pair satisfying the same conditions.

Example 19.2 Let κ be a fixed uncountable cardinal, and let K = κ + 1 be considered as a compact line. Let Γ consist of all closed countable sets A ⊆ κ such that 0 ∈ A and isolated points of A are isolated in κ. In particular, A can contain only ordinals of countable cofinality. Consider Γ as a σ -directed set endowed with inclusion. Given A ∈ Γ , define rA : K → K by setting rA(x) = max(A ∩[0,x]).It is easy to check that rA is a continuous retraction and, moreover, {rA}A∈Γ is a retractional skeleton in K.Ifκ ≥ℵ2, then this skeleton is not commutative. In fact, by Example 19.1(d), ℵ2 + 1 is not Valdivia compact and therefore cannot have a commutative retractional skeleton. Recall also a relevant result of Kalenda: The Banach space C(ℵ2 + 1) is not Plichko.

~~Given a retractional skeleton p ={rs}s∈Γ in a compact space K, denote D(p) = rs[K]. s∈Γ~~

~~As in the case of projectional skeletons, we shall call it the dense subspace induced by p. The following analogue of Theorem 17.10 holds.~~

~~Theorem 19.15 (Kubis[« 250]) Let p ={rs}s∈Γ be a retractional skeleton in a compact space K, and let D = D(p) be the dense subset induced by p. Then:~~

(a) D is countably tight and ℵ0-monolithic. (b) Disjoint relatively closed subsets of D have disjoint closures in K. In particular, D is FréchetÐUrysohn and K = βD. 436 19 Compact Spaces Generated by Retractions

Proof Part (a) is a consequence of Theorem 17.10 since p induces a projectional skeleton in C(K) that in turn induces a 1-norming subspace that contains D. For the proof of (b), ﬁx disjoint relatively closed sets A,B ⊆ D and suppose there is p ∈ A¯ ∩ B¯ , where the closure is taken in K. Fix a big enough cardinal θ and a countable M H(θ) so that A,B,p ∈ M.Letδ = sup(Γ ∩ M). Then rδ(p) ∈ D, and we may assume that rδ(p)∈ / A (interchanging the roles of A and B := [ ] ; t ; ∩ if necessary). Recall that Kδ rδ K is the limit of the inverse system Ks rs Γ t = [ ] ∈ ∩ M , where rs is the restriction of rs to Kt rt K . Thus, there are t Γ M and −1 an open set V ⊆ Kt such that U := Kδ ∩ rt [V ] is a neighborhood of p in Kδ −1 disjoint from A. On the other hand, rt [V ]∩A = ∅ because p is in the closure of A. Now note that Kt ∈ M is second-countable, and therefore M “knows” a ﬁxed countable open base of Kt . Thus, we may assume that V ∈ M. By elementarity, −1 there is a ∈ M ∩rt [V ]∩A. Finally, a ∈ Kδ (because D ∩M ⊆ Kδ), so a ∈ U ∩A, a contradiction.

~~Part (b) of the theorem above has a Banach space analogue that reads as follows.~~

Proposition 19.10 Let D be the norming space induced by a projectional skeleton ∗ {Ps}s∈Γ in a Banach space X. Then, for every weak continuous function f : D → R ∈ = ◦ ∗ there exists t Γ such that f f Pt D.

Proof Fix n>0, and consider Kn = nBX∗ , fn = f (D ∩ Kn). Then ={ ∗ } pn Ps Kn s∈Γ is a retractional skeleton in Kn, and Dn = D ∩ Kn is the dense subspace induced by pn. By Theorem 19.15, Kn is the CechÐStoneˇ compactiﬁcation of Dn, and therefore fn extends to a continuous map fn : Kn → R. Now, since pn determines a σ - directed inverse system of metric compact spaces, by Lemma 19.5 there is sn ∈ Γ = ◦ ∗ = ◦ ∗ such that fn f Ps Kn. In particular, fn f Ps Dn. We may assume that ≤ ≤ =n = ◦ ∗ n s1 s2 ....Lett supn∈ω sn. Then f f Pt D.

Theorem 19.16 Let K be a compact space, and let D be its dense subset. The following properties are equivalent: (a) There exists a retractional skeleton p on K such that D ⊆ D(p). (b) For every big enough cardinal θ, for every countable M (H (θ), ∈), the quotient map qM : K → K/M is one-to-one on (D ∩ M).

Proof The proof is very similar to that of Theorem 17.7, so we only sketch the arguments. (a) =⇒ (b): Let p ={rs}s∈Γ be a retractional skeleton such that D ⊆ D(p).Fix a countable M H(θ), where θ is big enough; assume p ∈ M. It is easy to verify M that the quotient map q induced by M equals rt , where t = sup(M ∩ Γ) (this is a general property of σ -continuous inverse systems of metric compact spaces). Finally, D ∩ M is contained in the range of rt . 19.10 Retractional skeletons 437

(b) =⇒ (a): Fix M H(θ) so that qM (D ∩ M) is one-to-one. Note that qM [D ∩ M] is dense in K/M. Indeed, a basic open set in K/M is of the form φ−1[U], where φ ∈ C(K) ∩ M and U is an open rational interval. The density of D and the elementarity of M together imply that f(d)∈ φ−1[U] for some d ∈ D ∩ M. It follows that qM is a retraction, and its right inverse j M is deﬁned as follows: j M (y) is the unique element of x ∈ (D ∩ M) such that qM (x) = y. Finally, the family of all qM ’s with suitable M H(θ) forms a retractional skeleton satisfying (a).

Corollary 19.19 (Bandlow [43]) Let K be a compact space. Then K is Corson if and only if, for every large enough cardinal θ, for every countable elementary substructure M of (H (θ), ∈) the quotient map qM : K → K/M is one-to-one on (K ∩ M).

~~A direct application gives the following stability result. It had been noticed ﬁrst by Michael and Rudin [292] and proved independently by Gul’ko [195].~~

~~Theorem 19.17 The class of Corson compact spaces is stable under continuous images.~~

Proof Fix a continuous surjection f : K → L with K Corson. Fix a countable M H(θ), where θ is big enough and f ∈ M.Fixp = q in (L ∩ M).Let −1 −1 KM = (K ∩ M), and let A = f (p) ∩ KM , B = f (q) ∩ KM . By the fact that K is Corson, the map rM : K → K defined by the conditions ∈ ∼ M rM (x) KM and rM (x) M x is homeomorphic to qK (it induces the equivalence relation ∼M ). Note that a ∼M b whenever a ∈ A and b ∈ B. Thus, by elementarity and compactness, there is φ ∈ C(K)∩M such that φ(a)<1/3fora ∈ A and φ(b)>2/3for b ∈ B. Indeed, fixing a ∈ A, for each b ∈ B there is a function φb ∈ C(K) ∩ M such that φb(a) < 1/3 and φb(b) > 2/3. Taking the minimum of finitely many φb’s for −1 +∞ ∈ ∩ which the sets φb (2/3, ) cover B, we obtain a single function ψa C(K) M such that ψa(a) < 1/3 and ψa(b) > 2/3 for every b ∈ B. A similar argument (now with the maximum of finitely many functions) gives the desired φ. Let C = φ−1(−∞, 1/3], D = φ−1[2/3, +∞). Note that we have identified M K/M with KM ,soφ = φ KM . Then C,D ∈ M are disjoint closed sets. Con- −1 −1 −1 −1 sequently, C = rM [C]=φ [C], D = rM [D]=φ [D] are closed and disjoint. Since they belong to M, there is in M a function ψ ∈ C(K) satisfying ψ C = 0 and ψ D = 1. It now remains to observe that f [C ] and f [D ] are disjoint. Indeed, if f(c ) = f(d ) for some c ∈ C , d ∈ D , then, setting c = rM (c ), d = rM (d ), we would have f(c)= f(c ) = f(d ) = f(d), but, on the other hand, c ∈ C, d ∈ D, which contradicts the fact that C ∩ D =∅. Finally, p ∈ f [C ], q ∈ f [D ], and these sets are in M. Therefore they are separated by a continuous function from M. 438 19 Compact Spaces Generated by Retractions

We shall also use Bandlow’s characterization for proving certain dichotomies concerning continuous images (see Theorem 19.22 below). Below we give an application to recognizing Corson compact spaces in dual Banach spaces.

~~Theorem 19.18 (Kubis[« 250]) Let X be a Banach space, and assume D ⊆ X generates projections in X. Then every weak∗ compact subset of D is Corson.~~

Proof Fix a weak∗ compact subset K ⊆ D.LetM H(θ) be countable, K,D ∈ M, and assume θ is big enough. Fix f = g in the weak∗ closure of K ∩ M. There is x ∈ X such that f(x)= g(x).LetPM be the projection generated by (X,D,M), = ∗ ∩ and let y PM x. Notice that PM is a projection onto the weak closure of D M. Therefore = = = f(y) fPM x (PM f)x f(x) and similarly g(y) = g(x). Using the continuity of both f and g and the fact that y ∈ (X ∩ M), ﬁnd z ∈ X ∩ M such that f(z)= g(z). Now the evaluation function δz ∈ C(K), deﬁned by δz(v) = v(z), is an element of M. We have shown that δz separates f from g,sof ∼M g.

~~19.11 Primarily Lindelöf spaces~~

In this section, we prove that, given a Valdivia compact K, the space C(K) endowed with the pointwise convergence on a Σ-subset has a property that is a strengthening of being Lindelöf. For Corson compact spaces, this result is due to Alster and Pol [4], and the generalization is due to Kalenda [218]. Recall that C(K) endowed with the topology of pointwise convergence on a dense Σ-subset is Lindelöf by Theorem 17.1. The Lindelöf property does not behave well with respect to products (see, for example, e.g. [146]). There is, however, a stronger property that is stable under countable products: being primarily Lindelöf. A space is primarily Lindelöf if it is a continuous image of a closed subspace of ℵ (Lκ ) 0 for some cardinal κ, where Lκ denotes the one-point Lindelöﬁcation of the discrete space κ. More precisely, Lκ = κ ∪{∞}, where all the points of κ are isolated while a neighborhood of ∞ is of the form U ∪{∞}, where κ \ U is countable. The following fact is obvious.

~~Lemma 19.12 The class of primarily Lindelöf spaces is stable under closed subspaces, continuous images, countable products and countable unions.~~

~~Lemma 19.13 Every primarily Lindelöf space is Lindelöf.~~

~~Proof It sufﬁces to show that for every ﬁxed uncountable cardinal κ the space L = ℵ (Lκ ) 0 is Lindelöf. The proof will be divided into two parts: 19.11 Primarily Lindelöf spaces 439~~

(1) For every open family U in L, there is a σ -disjoint open reﬁnement. (2) Every set A ⊆ L of cardinality ℵ1 has a complete accumulation point. Recall that V is a reﬁnement of U if V = U and for every V ∈ V there is U ∈ U with V ⊆ U. We now show that (1)+(2) implies thatL is Lindelöf. Fix an open cover U U = U U of L. By (1), we may assume that n∈ω n, where each n is a disjoint family. Then, for each p ∈ L, the family U (p) ={U ∈ U : p ∈ U } is countable. Suppose U has no countable subcover, and choose points aα ∈ L inductively so that aα ∈/ U (aξ ) for ξ<α. Since each U (aξ ) is countable, this is possible until we reach α = ω1.NowA ={aα : α<ω1} is a discrete set of cardinality ℵ1.Letp be its complete accumulation point. Choose V ∈ U such that p ∈ U . Then aα ∈ V for some α<ω1, and therefore aα+1 ∈/ V , a contradiction. n It remains to show (1) and (2). For the proof of (1), let Ln = (Lκ ) and observe that

~~(3) Every open family of subsets of Ln has a disjoint open reﬁnement.~~

This is evidently true for n = 1. Assuming (3) holds for Ln, note that Ln+1 = (Ln × κ) ∪ (Ln ×{∞}) and κ is discrete. Thus, given an open family U of subsets of Ln+1, first find a basic clopen neighborhood V = V0 × ...Vn of ∞,...,∞ that refines some element of U . Next, using the induction hypothesis, find a disjoint open family in (Ln × κ) \ V that is a refinement of U (Ln × κ) ={U ∩ (Ln × κ): U ∈ U }. Finally, add disjoint open sets in (Ln ×{∞}) that form a refinement of U (Ln ×{∞}). The resulting open family is a disjoint refinement of U . Now fix an open family U of L, and let πn : L → Ln denote the canonical projection. We may assume that U consists of basic open sets and therefore U = U U −1[ ] ⊆ n∈ω n, where each n consists of sets of the form πn U with U Ln open. Thus, by (3), for each n the family Un has a disjoint open refinement Vn. Finally, V U n∈ω n is a σ -disjoint open refinement of . It remains to show (2). Let pn : L → Lκ be the projection onto the nth coordinate. Fix a set A ⊆ L of cardinality ℵ1. Inductively construct uncountable sets A ⊇ A0 ⊇ A1 ⊇ ... and elements x0,x1,... in Lκ so that xn is a condensation point of pn[An] (i.e., U \ pn[An] is countable for every neighborhood U of xn). This is possible because of the structure of Lκ . Let x =x0,x1,...∈L. Fix a neighborhood U of x. We may assume that U = U0 × ...Un−1 × X × X × .... For each i

~~Theorem 19.19 (Kalenda [218]) Let K be a Valdivia compact space with a dense Σ-subset D. Then C(K), endowed with the topology of pointwise convergence on D, is primarily Lindelöf.~~

Proof Let τp(D) denote the topology of pointwise convergence on D. By Proposi- tion 19.9, (C(K) ,τp(D)) can be regarded as a closed subspace of (C(K ), τp(D )), 440 19 Compact Spaces Generated by Retractions where K is a zero-dimensional Valdivia compact space and D is its dense

Σ-subset. Thus, we may assume that K = K and D = D . Since C(K) = [− ] n∈ω C(K, n, n ) and the class of primarily Lindelöf spaces is closed under countable unions, it suffices to show that C(K,I) is primarily Lindelöf. Further- more, according to Proposition 17.2,asK is zero-dimensional, C(K,I) is an image of C(K,2ω) via a map of the form f → f ◦ φ. Such a map of course preserves the ω ℵ topology τp(D). Finally, C(K,2 ) is naturally homeomorphic to C(K,2) 0 and again this homeomorphism preserves τp(D). We have thus reduced our proof to showing that C(K,2) is primarily Lindelöf when endowed with τp(D). τ We now assume that K ⊆ 2 and D = Σ(τ)∩ K. Define θ : Lτ → C(K,2) by = ∈ ∞ = θ(ξ) prξ K for ξ τ and θ( ) 0, where prξ denotes the projection onto the ξth coordinate. It is standard to check that θ is continuous at ∞. The fact that elements of D have a countable support is used here. In this way, we have obtained a primarily Lindelöf subspace L = θ[Lτ ] of C(K,2) that separates the points of K. Elements of L can be identified with clopen subsets of K, and C(K,2) corresponds to the Boolean algebra generated by L. Thus C(K,2) is obtained by closing L under Boolean algebraic operations, which in terms of functions looks as follows: m(f, g) = min{f,g} (intersection) and = − = = c(f ) 1 f (complement). In other words, C(K,2) n∈ω Ln, where L0 L and Ln+1 = Ln ∪ m[Ln × Ln]∪c[Ln]. This shows that C(K,2) is primarily Lin- delöf.

~~Corollary 19.20 (AlsterÐPol [4], Pol [333]) Let K be a Corson compact space. Then Cp(K) is Lindelöf.~~

It is worth noting that Theorem 19.19 can be reversed. Namely, if D is a countably closed dense subset of a compact space K such that (C(K) ,τp(D)) is primarily Lindelöf, then K is Valdivia and D is a Σ-subset. This was proved by Kalenda [218] and, in the case D = K, by Pol [333].

~~19.12 Corson compact spaces and WLD spaces~~

We now discuss the relation between Corson compact spaces and weakly Lindelöf determined Banach spaces. Recall that, according to [13], a Banach space E is weakly Lindelöf determined (WLD, for short) if its dual unit ball BE endowed with the weak∗ topology is Corson compact.

Theorem 19.20 Given a Banach space E, the following properties are equivalent: (a) E is WLD. (b) E generates projections in E. (c) There exists an injective weak∗ continuous linear operator mapping E into a Σ-product of the real lines. 19.12 Corson compact spaces and WLD spaces 441

Proof (a) =⇒ (b): Assume K = BE is Corson, and fix a countable elementary submodel M of a big enough (H (θ), ∈) so that E ∈ M. Then K ∈ M. Using (the easier part of) Bandlow’s characterization, we deduce that the canonical quotient map qM : K → K/M is one-to-one on the weak∗ closure of K ∩ M. Thus we may as- ∗ sume that qM is a retraction onto K ∩ M . By Corollary 17.1, qM preserves the affine structure on K, and hence it is linear. More precisely, the map QM : E → E defined by QM (y) = y · qM (y/y) is linear. Now QM is a weak∗ continuous projection onto the weak∗ closed linear span of E ∩ M. Hence there exists a unique projection P M : E → E such that QM is the dual of P M . It remains to observe that im P M = E ∩ M and ker P M = ⊥ E ∩ M . (b) =⇒ (c): We use Theorem 17.15 and transfinite induction on the density of the space. Fix a Banach space E such that E generates projections, dens E = κ>ℵ0, and assume the implication (b) =⇒ (c) holds for Banach spaces of density <κ.Re- call that E has a projectional skeleton p such that E = D(p). By Theorem 17.6, E has a PRI {Pα}α<κ such that each im Pα has the same property as E, namely im Pα has a projectional skeleton that induces the full dual space. By Theorem 17.10(b), ∗ = E is weak countably tight, and therefore E α<κ PαE , unless κ has countable cofinality. In the first case, we conclude, using Theorem 17.15, that (E, E ) is a Plichko pair and hence (c) holds. In the latter case, also using Theorem 17.15, = we deduce that (E, D) is a Plichko pair, where D α<κ PαE . In particular, there exists a one-to-one weak∗ continuous linear operator T : E → RΓ such that T [D]⊆Σ(Γ). On the other hand, since κ has countable cofinality, necessarily TE ⊆ Σ(Γ). Thus E satisfies (c). Implication (c) =⇒ (a) is trivial.

~~Corollary 19.21 Every WLD Banach space has the controlled separable projection property.~~

Proof Let E be a WLD Banach space. By the theorem above, E generates projections in E and hence, given countable sets A ⊆ E, B ⊆ E , taking a countable elementary submodel M of a big enough H(χ) that A,B ⊆ M, we obtain a norm- one projection P : E → E such that A ⊆ im P and B ⊆ im P .

~~The following is a well-known characterization of WLD spaces of continuous functions ﬁrst proved by Argyros, Mercourakis and Negrepontis.~~

Theorem 19.21 (ArgyrosÐMercourakisÐNegrepontis [15]) Let K be a compact space. The following conditions are equivalent: (a) C(K) is WLD. (b) P(K)is Corson compact. (c) K is a Corson compact space such that P(K)is countably tight. 442 19 Compact Spaces Generated by Retractions

~~(d) K is Corson compact, and every positive regular Borel measure on K has a separable support.~~

Proof Implications (a) =⇒ (b) =⇒ (c) are trivial. (c) =⇒ (d): Fix a positive regular Borel measure μ on K. Normalizing μ,we may assume that it belongs to P(K).Nowμ belongs to the weak∗ closure of the convex hull of {δx : x ∈ K}, where δx is the probability measure supported by {x}. Applying (c), we get a countable set D ⊆ K such that ∗ μ ∈ conv{δx : x ∈ D} .

Clearly, the support of μ is contained in D¯ . ∗ =⇒ = (d) (a): Let B BC(K) be endowed with the weak topology. Then B is Valdivia compact by Corollary 19.18.AsK is Corson, its separable subsets are metrizable. Consequently, if {rs : K → Ks}s∈Σ is a retractional skeleton such that = = s∈Σ Ks K, then B s∈Σ BC(Ks ) is induced by a retractional skeleton and therefore is countably tight by Theorem 19.15(a). This shows that B is a Corson compact space.

It is interesting to note that, assuming the continuum hypothesis, there exist nonseparable Corson compact spaces carrying a strictly positive regular probability measure; see [15]. On the other hand, Martin’s axiom implies that every Corson compact satisfying the countable chain condition (i.e., every collection of open sets is countable) is separable. In particular, assuming Martin’s axiom, P(K) is Corson compact if and only if K is Corson if and only if C(K) is WLD.

~~19.13 A dichotomy~~

~~Below we prove a dichotomy that implies that countably tight members of class RC are Corson. For continuous images of Valdivia compact spaces, this is due to Kalenda [220].~~

~~Theorem 19.22 (Kubis[« 253]) Let K ∈ RC. Then either ω1 + 1 embeds into K or else K is Corson compact.~~

~~In the proof, we shall use Bandlow’s characterization (Corollary 19.19)interms of elementary submodels. First, we need a technical fact.~~

S = ; m; Lemma 19.14 Let Kn rn ω be an inverse sequence of internal retractions = S : → of compact spaces with K ←−lim , represented so that all projections rn K Kn are internal retractions. Let f : K → Y be a surjection and, for each n ∈ ω, let Bn be an open base for f [Kn] that is closed under ﬁnite unions. Then the family

~~−1 C ={Y \ f [K \ (f rn) [B]]: n ∈ ω, B ∈ Bn} 19.13 A dichotomy 443 is an open base for Y . In particular,~~

~~w(Y ) = sup w(f [Kn]). n∈ω~~

~~Proof Fix y ∈ Y and its neighborhood U.Findm ∈ ω and an open set W ⊆ Km such that −1 −1 −1 f (y) ⊆ rm [W]⊆f [U]. (19.6) Let −1 F = f [K \ rm [W]].~~

~~Then y/∈ F , and therefore U1 = Y \ F is a neighborhood of y, contained in U.Now ﬁnd n ≥ m and an open set W1 ⊆ Kn such that~~

~~−1 −1 −1 f (y) ⊆ rn [W1]⊆f [U1].~~

Thus −1 rn[f (y)]⊆W1 −1 −1 and W1 ⊆ f [U1]. The latter inclusion follows from the fact that W1 ⊆ rn [W1]. −1 Thus frn[f (y)]⊆U1. Using compactness and the fact that Bn is closed under ﬁnite unions, we can ﬁnd B ∈ Bn such that

~~−1 frn[f (y)]⊆B ⊆ U1.~~

~~We have −1 −1 −1 rn[f (y)]⊆f [B]⊆rm [W], −1 where the latter inclusion follows from the equality U1 = Y \f [K \rm [W]]. Thus~~

~~−1 −1 −1 −1 −1 −1 −1 f (y) ⊆ rn [f [B]] ⊆ rn [rm [W]] ⊆ (rmrn) [W]=rm [W] (19.7)~~

~~Hence, using (19.6) and (19.7), we obtain~~

~~−1 −1 −1 −1 f (y) ⊆ rn [f [B]] ⊆ f [U].~~

~~Finally, set −1 V = Y \ f [K \ (f rn) [B]]. Then V ∈ C and y ∈ V ⊆ U.~~

~~We are ready to prove Theorem 19.22.~~

Proof Every second-countable compact space is Corson, and therefore the dichotomy above holds for spaces of RC-rank 0. Fix an ordinal β>0, and assume 444 19 Compact Spaces Generated by Retractions that the statement above is true for spaces of RC-rank <β.FixX ∈ RC such that rkRC (X) = β.Let S = ; η; Kξ rξ κ be a continuous inverse system of internal retractions such that X = f [K] for some : → = S map f K X, where K ←−lim and rkRC (Kξ )<βfor every ξ<κ. We assume that κ is a regular cardinal. Let Xξ = f [Kξ ] for ξ<κ, and let Xκ = X.IfXξ contains a copy of ω1 + 1forsomeξ<κ, then there is nothing to prove, so assume that Xξ is Corson compact for each ξ<κ. Suppose ﬁrst that κ>ℵ0 and X = Xξ . (*) ξ<κ = ∈ ∈ Then K ξ<κ Kξ . Fix y K so that f(y)/ ξ<κ Xξ . Recall that

~~y = lim rξ (y). ξ→κ~~

~~Also,~~

rδ(y) = lim rξ (y) ξ<δ by the continuity of the sequence S. Thus the map φ : κ +1 → K, deﬁned by φ(ξ)= frξ (y) and φ(κ) = f(y), is continuous. By assumption, φ(ξ) = f(y) for every ξ<κ, and therefore the sequence

{φ(ξ)}ξ<κ does not stabilize. Find a closed coﬁnal set C ⊆ κ such that φ C is one-to-one. Then Z = φ[C ∪{κ}] ⊆ X is homeomorphic to the linearly ordered space κ + 1. Since κ>ℵ0, this shows that X contains ω1 + 1. Now suppose that (*) does not hold. We claim that X is Corson. For this, we use Bandlow’s characterization. Fix a big enough cardinal χ and a countable submodel M H(χ) such that f,S ∈ M. Let δ = sup(κ ∩ M)

(if κ =ℵ0, then of course δ = ω ⊆ M and Xδ = X). Define M = ∩ M = ∩ Xξ (Xξ M), Kξ (Kξ M), for ξ<κ.LetqM : X → X/M denote the quotient map induced by M. In order to M M show that qM is one-to-one on X := (X ∩ M), it suffices to find a base for X that is contained in M. 19.13 A dichotomy 445

~~Notice that~~

X ∩ M ⊆ Xδ. =ℵ = = Indeed, if κ 0, then Xδ X; otherwise, X ξ<κ Xξ and therefore, by elementarity, X ∩ M ⊆ Xξ ⊆ Xδ. ξ∈κ∩M Now observe that [ M ]= M f Kξ Xξ for every ξ ∈ κ ∩ M. Indeed, if x ∈ Xξ ∩ M and ξ ∈ M, then, by elementarity, x = f(y)for some y ∈ Kξ ∩ M, which shows that

~~Xξ ∩ M ⊆ f [Kξ ∩ M].~~

~~Clearly,~~

~~f [Kξ ∩ M]⊆Xξ ∩ M. Note that f [KM ]=XM because f ∈ M. On the other hand, M = M = M K Kδ Kξ . ξ∈κ∩M~~

~~Let~~

{ξn : n ∈ ω}⊆δ ∩ M = be an increasing sequence such that δ supn∈ω ξn.Let S = ; m; 0 Zn pn ω , where Z = KM and p = r Z . We apply Lemma 19.14 to ﬁnd a suitable base n ξn n ξn n M ∈ ∈ for X .Fixn ω. Then Xξn M, and by assumption it is Corson compact. Thus M induces a quotient map that is one-to-one on XM . In particular, the family ξn B ={ ∈ : } n U M U open in Xξn forms a base for XM more precisely, U ∩XM : U ∈ B is an open base for XM . ξn ξn n ξn Clearly, Bn is closed under ﬁnite unions. Lemma 19.14 says that

~~M M −1 C ={X \ f [K \ (fpn) [B]]: n ∈ ω, B ∈ Bn} is an open base for XM = f [KM ].Let~~

~~−1 C ={X \ f [K \ (fpn) [B]]: n ∈ ω, B ∈ Bn}. 446 19 Compact Spaces Generated by Retractions~~

~~Then C ⊆ M and {C ∩ XM : C ∈ C }=C . Thus we have shown that M contains a family that forms an open base for XM . Since M was arbitrary, this implies that X is Corson compact.~~

~~19.14 Alexandrov duplications~~

We are going to show that, contrary to Valdivia compact spaces, class R contains some nonmetrizable compactiﬁcations of the natural numbers. Therefore R contains separable spaces that are not continuous images of Valdivia compact spaces. Fix a topological space X and its subset A. Formally, assume X ∩ (X × 2) =∅. Deﬁne the space Dbl(X, A) as

~~(X ×{0}) ∪ (A ×{1}), endowed with the topology that each point of the form (x, 1) is isolated, and a neighborhood of (x, 0) is of the form~~

~~(U ×{0, 1}) \{(x, 1)}, where U is a neighborhood of x in X. This construction is due to Alexandrov and is sometimes called the Alexandrov duplication of A in X.~~

~~Proposition 19.11 Dbl(K, A) ∈ R whenever K ∈ R and A ⊆ K. Moreover, if A is countable, then rkR(Dbl(K, A)) ≤ rkR(K) + 1.~~

~~Proof Fix K ∈ R and A ⊆ K. We use induction on the cardinality of A. Clearly, Dbl(K, A) ∈ R if A is ﬁnite, so assume~~

~~|A|=κ ≥ℵ0 and Dbl(K, B) ∈ R whenever B ∈[K]<κ . Write~~

~~A ={aξ : ξ<κ}.~~

~~Deﬁne Kξ = Dbl(K, Aξ ), where Aξ ={aη : η<ξ}, and let~~

~~rξ : Dbl(K, A) → Kξ be deﬁned by rξ (x) = x for~~

~~x ∈ (K × 0) ∪ (Aξ × 1) and rξ (x, 1) = (x, 0) for x ∈ A \ Aξ . Clearly, rξ is a retraction and, by the inductive hypothesis, Kξ ∈ R for every ξ<κ.Wealsohave~~

rξ ◦ rη = rξ = rη ◦ rξ 19.14 Alexandrov duplications 447 whenever ξ<η. Finally, observe that for every x ∈ Dbl(K, A) we have rξ (x) = x for all but boundedly many ξ<κ.Thus,byLemma19.4,wehave = ; η; Dbl(K, A) ←−lim Kξ rξ κ , η = where rξ rξ Kη. The second statement follows from the proof above and from the fact that

~~rkR(K ⊕ S) ≤ rkR(K) whenever |S| < ℵ0 (see Corollary 19.8).~~

ℵ Note that Dbl(2 1 ,D) is a nonmetrizable compactification of the natural num- ℵ bers whenever D ⊆ 2 1 is countable and dense. Thus, the class R contains nonmetrizable compactifications of ω. We finish this section by exploring more proper- ℵ ties of the space Dbl(2 1 ,D).

~~Proposition 19.12 Assume K is a compact space with a countable dense set consisting of Gδ points. If w(K) has uncountable coﬁnality, then C(K) has no PRI.~~

~~Proof Let λ = w(K), and suppose~~

{Pα : α<κ} is a PRI on C(K). Taking a coﬁnal subsequence, we may assume that κ = cf(λ). ℵ Then κ> 0.LetPα denote the adjoint operator on C(K) .Wehave = = lim (Pξ μ)f lim μ(Pξ f) μ(f ), ξ→κ ξ→κ ∗ = ∈ which implies that weak limξ→κ Pξ μ μ for every μ C(K) . Moreover, each ∗ ∗ Pα is (weak ,weak )-continuous. Denote by B the closed unit ball of C(K) . Then { : } PαB α<κ ∗ is an increasing sequence of weak closed subsets of B (we have PαB ⊆ B because Pα=1). Moreover,

PαB α<κ is weak∗ dense in B. Now assume that x ∈ K is a Gδ point, and ﬁx functions fn : K →[0, 1] such that −1 fn [(0, 1]] = {x}. n∈ω 448 19 Compact Spaces Generated by Retractions

~~Then~~

~~{δx}={μ ∈ B : (∀ n ∈ ω) μ(fn) = 1}, ∗ and the set on the right-hand side of the equality above is Gδ in (B, weak ) (δx denotes the Dirac measure of x). It follows that~~

~~∈ { : ∈ } δx Pαδx α Cx~~

(the closure in σ(X X)) for some countable set Cx ⊆ κ. Since cf(κ) > ℵ0, it follows that ∈ δx Pβ(x)B for some β(x) < κ. Now let D ⊆ K be a countable dense set consisting of Gδ points. Then { : ∈ }⊆ δx x D Pβ B, where β = sup β(x) < κ. x∈D ∗ It remains to observe that Pβ C(K) is a proper weak closed linear subspace of C(K) , while at the same time the set

~~{δx : x ∈ D} is linearly weak∗ dense in C(K) . This is a contradiction.~~

~~ℵ ℵ Theorem 19.23 Let K = Dbl(2 1 ,D), where D ⊆ 2 1 is countable and dense. Then~~

~~(i) K ∈ R and rkR(K) = 2; (ii) K is not a continuous image of a Valdivia compact space; (iii) C(K) has no PRI; and ℵ (iv) C(K) is isomorphic to C 2 1 .~~

Proof Part (i) follows from the second part of Proposition 19.11. Part (ii) is a consequence of the fact that continuous images of Valdivia compact spaces have dense monolithic sets. Part (iii) follows from Proposition 19.12.Part(iv)isprovedbya standard decomposition method; see, for example, [50].

~~19.15 Valdivia compact groups~~

We show that every Valdivia compact group is homeomorphic to a product of metric compact spaces. The case of Abelian groups (in fact, a stronger statement) has 19.15 Valdivia compact groups 449 already been presented in Theorem 19.5 and comes from [251]. The general case has been proved by Chigogidze [95] using different arguments. The following lemma is crucial.

Lemma 19.15 Let f : G → H be a continuous epimorphism of topological groups, and assume there exists a continuous map j : H → G such that f ◦ j = idH . Then there exists a homeomorphism θ : G → H × ker f such that

~~θ G H × ker f~~

~~f prH H~~

~~= ◦ : × → (i.e., f prH θ, where prH H ker f H is the projection).~~

~~Note that θ is not necessarily a group homomorphism.~~

~~Proof Let K = ker f , and deﬁne θ : G → H × K by setting~~

~~θ(x)= (f (x), j ◦ f(x)· x−1).~~

It is clear that θ is continuous and well deﬁned since − − − f j ◦ f(x)· x 1 = f ◦ j ◦ f(x)· f(x) 1 = f(x)· f(x) 1 = 1, where 1 denotes the unit of G. Now deﬁne ρ : H × K → G by

~~ρ(p,q)= q−1 · j(p).~~

~~Clearly, ρ is continuous. Given x ∈ G and (p, q) ∈ H × K,wehave~~

~~ρ(θ(x))= ρ(f (x), j (f (x)) · x−1) =[j(f(x))· x−1]−1 · j(f(x))= x and~~

~~θ(ρ(p,q))= (f (q−1 · j (p)), j ◦ f(q−1 · j(p))·[q−1 · j(p)]−1) = (f (j (p)), j ◦ f ◦ j(p)· j(p)−1 · q) = (p, q).~~

~~Thus ρ ◦ θ = idG and θ ◦ ρ = idH ×K , which shows that θ is a homeomorphism. ◦ = Finally, prH θ f .~~

S = ; η; Lemma 19.16 Let Xξ pξ κ be a continuous inverse sequence of topological spaces, and assume that for each α<κthere are a space Yα and a homeomorphism 450 19 Compact Spaces Generated by Retractions

~~+ : + → × α 1 = ◦ θα Xα 1 Xα Yα such that pα prXα θα, where prXα is the projection. S Then ←−lim is homeomorphic to X0 × Yα. α<κ~~

= × ; η; Proof Let Zα X0 ξ<α Yξ , and let Zξ prξ κ be the natural continuous inverse sequence of projections. Its limit is Zκ . Inductively deﬁne homeomorphisms τα : Xα → Zα that commute with both inverse sequences. The limit step is straightforward because of the continuity of both sequences. Suppose τα has been deﬁned, and let

~~= × ◦ τα+1 (τα idYα ) θα.~~

~~Then τα+1 is a homeomorphism for which the following diagram commutes.~~

~~α+1 pα Xα Xα+1~~

~~τα τα+1 α+1 prα Zα Zα+1~~

~~Finally, the limit of τα’s is the desired homeomorphism.~~

~~Theorem 19.24 (Chigogidze [95]) Every topological group that is a topological retract of a Valdivia compact space is homeomorphic to a product of metric compact groups.~~

Proof We use induction on the weight of G. The statement above is obvious in the case w(G) ≤ℵ0. Assume κ = w(G) and the theorem is true for groups of weight <κ that are retracts of Valdivia compact spaces. Fix a group G with w(G) = κ and fix a retraction r : K → G, where K is a = { } Valdivia compact space. Let λ cf κ, and fix a continuous chain Mα α<λ of el- ∈ ementary substructures of a given H(χ) big enough that r M0 and α<λ Mα contains a fixed open base of G.LetGα = G/Mα, and let pα be the canonical quotient of G onto Gα. By Corollary 17.1, Gα is a topological group and pα is a group homomorphism. In this way, we obtain an inverse sequence of compact S = ; η; η groups Gξ pξ κ whose limit is G and each pξ is a continuous epimorphism. 19.16 Compact lines in class R 451

~~We claim that each pα has a topological right inverse. Fix α, and let qα : K → K/Mα be the canonical quotient. We have the following commutative diagram.~~

~~r K G~~

~~qα pα~~

~~K/Mα G/Mα rMα~~

By Lemma 17.6, qα is a retraction. In order to conclude that pα is right-invertible, it sufﬁces to show that rMα is. For convenience, assume that r is an internal retraction (i.e., G ⊆ K and r G = idG). Let Z = qα[G]⊆K/Mα.Giveny ∈ G,wehavepα(y) = pα(r(y)) = M M M r α (qα(y)). Thus, r α Z is one-to-one and r α [Z]=G/Mα, which shows that rMα is a retraction. Finally, Lemmas 19.15 and 19.16 show that G is homeomorphic to the product G0 × Hα, α<λ

= α+1 × where Hα ker(pα ) for α<λ. In particular, Gα+1 is homeomorphic to Gα Hα, and hence Hα is a retraction of Gα+1. By the induction hypothesis, each Hα is homeomorphic to a product of metric compact groups.

~~The following question is left open: Is every compact group from class R homeomorphic to a product of metric compact spaces?~~

~~19.16 Compact lines in class R~~

In this section, we come back to the discussion of compact lines in the classes R and RC and their relations to Valdivia compact spaces. Recall that, by Nakhmanson’s theorem (Theorem 18.1) and Pol’s theorem (Corollary 19.20), Corson compact lines are necessarily metrizable.

~~Theorem 19.25 ([253]) Valdivia compact lines have weight ≤ℵ1.~~

Proof Let K be a Valdivia compact line. An argument similar to the one showing that ω2 + 1 is not Valdivia proves that the character of K is at most ℵ1. The rest of the proof requires an argument from logic: Supposing that K has weight > ℵ1, one can construct a model of set theory in which there exists a nonmetrizable linearly ordered Corson compact. We refer to [253] for the details.

~~Theorem 19.26 Linearly ordered spaces in class RC are monolithic. 452 19 Compact Spaces Generated by Retractions~~

~~Before proving the statement above, we need the following fact that is a consequence of the dichotomy from Theorem 19.22 and Nakhmanson’s theorem.~~

~~Lemma 19.17 Assume X ∈ RC is linearly ordered and f : X →[0, 1] is an increasing surjection. Then the set S(f) ={t ∈[0, 1]: |f −1(t)| > 1} is countable.~~

~~Proof Suppose S(f) is uncountable. Deﬁne ⎧ ⎨[0, 1] if t ∈ S(f), f −1(t) is connected, = { } −1 Y(t) ⎩ 0, 1 if f (t) is disconnected, {0} otherwise.~~

For each t ∈[0, 1],let −1 ht : f (t) → Y(t) be an order-preserving surjection. Let Y be the lexicographic product of {Y(t)}t∈[0,1]. Then the family {ht }t∈[0,1] induces an order-preserving surjection h: X → Y . Since S(f) is uncountable, Y ∈ RC is a nonmetrizable ﬁrst-countable space. This contradicts Theorem 19.22 because linearly ordered Corson compact spaces are metrizable.

~~We are ready to prove Theorem 19.26.~~

~~Proof Let X ∈ RC be a linearly ordered compact space. Fix a countable set D ⊆ X, and let Y = D. Deﬁne an equivalence relation ∼ on X as follows:~~

~~x ∼ y ⇐⇒ [ x,y]∩Y is scattered.~~

Note that the closure of a scattered set in a linearly ordered compact space is scattered. Observe that the ∼-equivalence classes are closed and convex. Let K = X/ ∼ and let q : X → K be the quotient map. Then K is a connected and separable linearly ordered space and therefore order-isomorphic to [0, 1]. By Lemma 19.17,the set − S(q) ={t ∈ K :|q 1(t)| > 1} is countable. For each t ∈ S(q), choose a countable family Bt of open intervals of X that separates the points of q−1(t) ∩ Y (recall that q−1(t) ∩ Y is scattered, by the deﬁnition of ∼). Let −1 B ={q [(r, →)]: r ∈ Q}∪ Bt , t∈S(q) where Q ⊆ K is a countable dense set. Then B is a countable family of open sets that separates the points of Y . Thus w(Y ) ≤ℵ0.

~~We now turn to connected spaces in class R. 19.16 Compact lines in class R 453~~

→ Recall that the long ray is the linearly ordered space R = ω1 ·[0, 1) (where P · Q denotes the lexicographic product of P with Q). Note that R→ + 1isa connected linearly ordered compact space with exactly one point of uncountable character (namely, the maximal point). Moreover, each proper interval that does not contain the maximal point is separable. Denote by ←R→ the unique linearly ordered space K =[a,b] such that for every internal point p ∈ (a, b) the interval [p,b] is order-isomorphic to R→ + 1 and the interval [a,p] is order-isomorphic to the inverse of R→ + 1.

~~Proposition 19.13 The spaces R→ + 1 and ←R→ are Valdivia compact.~~

~~Proof Let Kα = δ ·[0, 1) + 1, and deﬁne → rα : R + 1 → Kα by setting rα(x) = x if x ∈ Kα and rα(x) = max(Kα) otherwise. Then~~

~~{rα : α<ω1} → is an inverse sequence of internal retractions in R + 1 and each Kα is second- countable. This shows that R→ + 1 is Valdivia compact. The case of ←R→ is similar.~~

~~Theorem 19.27 (Kubis[« 253]) Let K ∈ R be a nonempty connected linearly ordered space. Then K is order-isomorphic to one of the following spaces:~~

~~→ → − ← → 1, [0, 1],R + 1,(R + 1) 1, R .~~

~~In particular, K is Valdivia compact.~~

~~Proof Suppose the statement is false, and let K be a counterexample of a minimal possible R-rank. Fix an inverse sequence of internal retractions~~

~~{rα : α<κ} in K such that rkR(Kα)~~

~~{bα}α<κ is strictly increasing. Pick p ∈ (a0,b0). By the minimality of rkR(K), each of the intervals [aα,bα] is order-isomorphic to one of the spaces listed in the theorem. It follows that~~

[p,bα]⊆(aα+1,bα+1) 454 19 Compact Spaces Generated by Retractions is order-isomorphic to [0, 1]. Hence, κ ∈{ℵ0, ℵ1}.LetK =[a,b]. Then [p,b] is → order-isomorphic either to [0, 1] (if κ =ℵ0)orR + 1(ifκ =ℵ1). By the same argument, [a,p] is order-isomorphic either to [0, 1] or the reversed R→ + 1. This shows that K is not a counterexample, a contradiction.

We now turn to Valdivia compact lines. As we have already proved, all of them have weight ≤ℵ1. Below we construct a universal one, namely a Valdivia compact line that maps onto any other (nonempty!) Valdivia compact line. Let 2ω denote, as usual, the Cantor set. Denote by π : 2ω → 2ω the unique increasing map satisfying the following conditions. (1) π −1(t) is order-isomorphic to the Cantor set whenever t is external in 2ω (i.e., t is isolated from one side (in particular, when t = 0ort = 1)). (2) |π −1(t)|=1ift is internal in 2ω. It is easily seen that π is a retraction. A possible right inverse i : 2ω → 2ω to π can be deﬁned by setting i(p) = min π −1(p) in the case where p is isolated from the right, i(p) = max π −1(p) in the case where p is isolated from the left, and i(p) = p otherwise. Given two increasing maps f,g : X → Y , we shall say that f is order-isomorphic to g if there are order isomorphisms θX : X → X and θY : Y → Y such that θY ◦f = g ◦ θX.

Lemma 19.18 (a) π ◦ π is order-isomorphic to π. S = ; n ; = ω n+1 = ∈ S (b) Let Cm pm ω , where Cn 2 and pn π for each n ω. Then ←−lim ω S is order-isomorphic to 2 and every projection from ←−lim is order-isomorphic to π.

Proof It is clear that π ◦π satisﬁes conditions (1) and (2) above. For the proof of (b), = S notice that C ←−lim is a compact zero-dimensional metric line without isolated points and therefore is order-isomorphic to 2ω. Finally, every projection from the limit again satisﬁes (1) and (2).

~~The construction is based on the following crucial fact.~~

Lemma 19.19 Let K and L be compact metric lines, and let f : 2ω → L and r : K → L be continuous increasing quotients. Assume further that r is a retraction. Then there exists a continuous increasing quotient g : 2ω → K for which the following diagram commutes.

~~g 2ω K~~

~~π r~~

~~2ω L f 19.16 Compact lines in class R 455~~

Proof We assume that L ⊆ K and r L = idL.LetQ denote the set of all external points of the Cantor set 2ω. −1 −1 Given z ∈ L, define Fz = π [f (z)]. Note that Fz is a closed interval and ω hence, since 2 is dense in itself, Fz is either order-isomorphic to the Cantor set or else |Fz|≤2. Now observe that the required map g must satisfy −1 g[Fz]=r (z) for every z ∈ L. (19.8) Conversely, if g : 2ω → K satisfies (19.8), then it is an order-preserving surjection such that r ◦ g = f ◦ π holds. Thus, it suffices to show that Fz has an order- preserving map onto r−1(z) for every z ∈ L. Fix z ∈ L. Then r−1(z) is a closed, metrizable interval in K containing z (recall that r is an internal retraction). If r−1(z) ={z}, then there is nothing to prove, so assume r−1(z) =[a,b], where a

Theorem 19.28 (Kubis[« 252]) The space Cω1 is a Valdivia compact line such that for every nonempty Valdivia compact line K there exists a continuous increasing surjection : → F Cω1 K. S Proof That Cω1 is Valdivia compact follows from the fact that is a retractive inverse sequence of metric compact spaces. Fix a nonempty Valdivia compact line K. By Theorem 19.25, the weight of K ℵ T = ; η; does not exceed 1.Let Kξ rξ ω1 be a continuous inverse sequence of continuous increasing maps onto compact metric lines. We may assume that |K0|=1. η Since K is Valdivia, we may also assume that each rξ is a retraction. ω β We construct inductively continuous surjections fα : 2 → Kα so that rα ◦ fβ = ◦ β { } fα πα holds for every α<β<ω1. The limit of fα α<ω1 will then be the desired increasing surjection from Cω1 onto K. Fix α<ω1, and suppose fξ have already been deﬁned for ξ<α.Ifα is a limit ordinal, then we deﬁne fα to be the limit of {fξ }ξ<α. The continuity of both sequences guarantees that fα is a continuous increasing surjection. Suppose now that α = ξ + 1. By Lemma 19.19, there exists a continuous increas- : ω → ξ+1 ◦ = ◦ = ◦ ξ+1 := ing surjection g 2 Kα such that rξ g fξ π fξ πξ . Set fα g. α ◦ = ◦ α By the induction hypothesis, rξ fα fξ πξ holds for every ξ<α. 456 19 Compact Spaces Generated by Retractions

~~We ﬁnish with an example answering the problem on subspaces of Plichko spaces.~~

~~Example 19.3 Let Kω1 be the connectiﬁcation of Cω1 (see Proposition 18.3). Then = E C(Kω1 ) is linearly isometric to a subspace of C(Cω1 ). On the other hand, E is not a Plichko space.~~

Proof Take a countable elementary submodel of some H(χ) big enough that E ∈ ∩ : → M. Then (E M) is induced by the quotient map qM Cω1 Cω1 /M. Observe that no interval of Cω1 is metrizable. It follows that the set

~~{ ∈ :| −1 | } t Cω1 /M qM (t) > 1 is everywhere dense. By Theorem 17.4, qM admits no averaging operator; that is C(K/M) is not complemented in E. This shows that E is not Plichko.~~

~~Notice that E has the regular separable complementation property by Theo- rem 18.3 together with Theorem 19.26.~~

~~19.17 More on Eberlein compact spaces~~

Recall again that a space K is called Eberlein compact if it is homeomorphic to a weakly compact subset of a Banach space. We recall also the following classical result of Amir and Lindenstrauss [5].

~~Theorem 19.29 (AmirÐLindenstrauss [5]) Eberlein compact spaces are precisely those compact spaces that embed topologically into (c0(Γ ), τp) for some set Γ .~~

Proof Fix a set Γ . We check that the pointwise topology coincides with the weak one on every bounded subset of c0(Γ ). For this, fix a bounded net {xs}s∈Σ ⊆ c0(Γ ) pointwise converging to zero. We shall show that it converges weakly to zero. We may assume that xs ≤ 1 for every s ∈ Σ.Fixy ∈ 1(Γ ) = c0(Γ ) and fix ε>0. There is a finite set A ⊆ Γ such that |y(i)| <ε/2. i∈Γ \A

~~Let n =|A|, and for each a ∈ A ﬁnd ta ∈ Σ such that~~

|xs(a)| <ε/(2n|y(a)|) whenever s ≥ ta.Lett ∈ Σ be such that ta ≤ t for a ∈ A.Ifs ≥ t, then 19.17 More on Eberlein compact spaces 457 | |= ≤ | |·| |+ | |·| | y(xs) xs(i)y(i) xs(a) y(a) xs(i) y(i) i∈Γ a∈A i∈Γ \A <ε/2 + |y(i)| <ε. i∈Γ \A

Hence |y(xs)| <εfor every s ≥ t. Since y was arbitrary, this shows that xs converges weakly to zero. We conclude that the pointwise and weak topologies coincide on compact subsets of c0(Γ ). In particular, every compact subset of (c0(Γ ), τp) is Eberlein compact. Now ﬁx an Eberlein compact space K inside a Banach space E with the weak topology. Without loss of generality, we may assume that K is linearly dense in E. Now E is weakly compactly generated and in particular has a projectional skeleton by Proposition 17.4 and Theorem 17.7. Hence, by Corollary 17.5, there is an injective bounded linear operator T : E → c0(Γ ) for some set Γ . It is clear that T is continuous with respect to the weak topologies on both spaces, and therefore K embeds into c0(Γ ) with the weak topology. By the remarks above, we can replace the weak topology by the pointwise one.

~~Corollary 19.22 Eberlein compact spaces are Corson.~~

As another corollary, we obtain the well-known covering-type characterization of Eberlein compact spaces originally due to Rosenthal. Recall that a family A is T0-separating in X if for every x,y ∈ X there is A ∈ A such that |{x,y}∩A|=1.

Theorem 19.30 (Rosenthal) Let K be a compact space. Then K is Eberlein if and {F } only if there exist a sequence n n∈ω of families of open Fσ subsets of K such that F F each n is point-ﬁnite and n∈ω n is T0-separating in K. If K is a zero-dimensional Eberlein compact space, then the sequence {Fn}n∈ω above may consist of clopen sets.

Proof Assume first that K has the sequence of families {Fn}n∈ω satisfying the conditions above. For each U ∈ Fn, choose a continuous map 1 f : K → 0, U n such that − 1 U = f 1 0, U n F = F (here we use the fact that U is open and Fσ ). Let n∈ω n and, finally, let F f : K →[0, 1] be the diagonal of {fU }U∈F (i.e., f (x)(U) = fU (x) for x ∈ K, U ∈ F ). Since F is T0-separating, f is injective. By compactness, it is a topological embedding. It remains to check that f [K]⊆c0(F ).Fixx ∈ K,fixε>0 458 19 Compact Spaces Generated by Retractions ∈ 1 | | ∈ F and choose k ω so that k <ε. Then fU (x) <εwhenever U n≥k n.Onthe other hand,

F0 ∪···∪Fk−1 is point-finite, and therefore |fU (x)| <ε for all but finitely many U ∈ F .This shows that f(x)∈ c0(F ). Assume now that K is an Eberlein compact space embedded into c0(Γ ) with the pointwise topology. Let {qn}n∈ω be a fixed enumeration of all nonzero rational numbers. Given n ∈ ω, define

~~Vα,n ={x ∈ K : x(α) > qn} if qn > 0 and~~

~~Vα,n ={x ∈ K : x(α) < qn} if qn < 0. Note that Vα,n is an open Fσ -set and for each x ∈ K,foraﬁxedn, there exist only ﬁnitely many α ∈ Γ such that x ∈ Vα,n. It follows that~~

Fn ={Vα,n : α ∈ Γ } F = is point-finite. It remains to check that n∈ω n is T0-separating. Fix x y in K and fix α such that x(α) = y(α). Suppose x(α) < y(α).Findn such that x(α) < qn 0, then y ∈ Vα,n and x/∈ Vα,n, while if qn < 0, then x ∈ Vα,n and y/∈ Vα,n. Assume now that K is a zero-dimensional Eberlein compact space and let {Fn}n∈ω be as above. Using zero-dimensionality and the fact that Vα,k is open Fσ , n we can find clopen sets Vα,k such that = n Vα,k Vα,k. n∈ω Now let F n ={ n : ∈ } k Vα,k α Γ . F n Then k is point-finite and F n = F k n n,k∈ω n∈ω F n is T0-separating. Arranging the families k into a sequence completes the proof of the second statement.

~~We now turn to the properties of spaces of continuous functions over Eberlein compact spaces.~~

Lemma 19.20 Let K be an Eberlein compact space. Then Cp(K) contains a set homeomorphic to the one-point compactiﬁcation of a discrete space that separates the points of K. 19.17 More on Eberlein compact spaces 459

Proof By Theorem 19.29,wehaveK ⊆ c0(κ) for some set κ, where c0(κ) is considered with the pointwise topology. Let fα : K → R be the restriction to K of the projection onto the αth coordinate. That is, fα(x) = x(α) for x ∈ K.Let

~~L ={fα : α ∈ κ}∪{0}.~~

Then L ⊆ Cp(K), and it separates the points of K. Fix a subbasic neighborhood V of 0 ∈ L. That is, V ={f :|f(x)| <ε} for some x ∈ K and ε>0. There is a ﬁnite set S ⊆ κ such that |x(α)| <ε for α ∈ κ \ S. Thus fα ∈ V if and only if

~~α ∈ κ \ S.~~

~~This shows that L is homeomorphic to the Alexandrov compactiﬁcation of the discrete set {fα : α ∈ κ}.~~

Because of the lemma above, it is natural to ask which compact spaces K have the property that some compact set L ⊆ Cp(K) that separates the points of K. It turns out that these are again Eberlein compact spaces. We have the following theorem.

~~Theorem 19.31 Let K be a compact space. Then K is Eberlein if and only if there exists a compact space L ⊆ Cp(K) that separates the points of K. In that case, L is also Eberlein compact.~~

~~Before proving this result, we need two lemmas that may be of independent interest.~~

~~Lemma 19.21 Let K and L be compact spaces such that L ⊆ Cp(K) and L separates the points of K. Then there exists K ⊆ Cp(L), homeomorphic to K, which separates the points of L.~~

Proof Given x ∈ K, deﬁne x(f) = f(x)for f ∈ L. Then x is a continuous function on L (i.e., x ∈ Cp(L)). It is clear that the map x → x is continuous. It is one-to-one because L separates the points of K.LetK ={x : x ∈ K}.Iff = g in L, then f(x)= g(x) for some x ∈ K and hence x(f) = x(g). This shows that K separates the points of L.

Lemma 19.22 Let K be a compact space, and let L be a compact subset of Cp(K) that separates the points of K. Furthermore, let M be an elementary submodel of some (H (χ), ∈) big enough that K,L ∈ M. Let K1 = (K ∩ M) and τp L1 = (L ∩ M) , where τp denotes the pointwise convergence topology. Then:

~~(a) There exists a unique retraction r : K → K that satisﬁes r[K]=K1 and f(r(x))= f(x)for every x ∈ K and for every f ∈ L1. 460 19 Compact Spaces Generated by Retractions~~

(b) The adjoint operator P : C(K) → C(K) (that is, the operator deﬁned by Pf = f ◦ r) is (τp,τp)-continuous and P [L]=L1. (c) L1 K1 separates the points of K1 and is compact with respect to τp on C(K1).

~~Lemma 19.22 says in particular that (K ∩ M) is homeomorphic to K/M and τ (L ∩ M) p is homeomorphic to L/M.~~

Proof Given S ⊆ C(K), deﬁne φS : K → C(S) by setting φS(x)(f ) = f(x). Then φS is continuous with respect to the pointwise topology on C(S). Given T ⊆ K, we denote by πT the restriction operator from C(K) to C(T ).Let = ∩ = ∩ [ ] K0 K M, L0 L M. We ﬁrst note that, by elementarity, φL0 K0 is dense in [ ] [ ] [ ] φL0 K and πK0 L0 is dense in πK0 L . Indeed, a basic neighborhood of φL0 (x) is of the form

~~L0 V ={p ∈ R : (∀ i~~

Proof (Proof of Theorem 19.31) By Lemma 19.21, the assumptions are symmetric, and therefore it suffices to show that L is Eberlein. In view of Theorem 19.29,it is sufficient to show that C(K) has a one-to-one bounded linear operator into some c0(Γ ). Using Lemma 19.22, we construct a PRI in C(K) onto subspaces of the form C(Kα), where Kα is a compact space of smaller weight satisfying the same assumptions as K. Specifically, letting κ be the weight of K, we fix a chain {Mα}α<κ 19.17 More on Eberlein compact spaces 461 of elementary submodels of H(χ) big enough that |Mα|≤α +ℵ0, K,L ∈ M0 and α<κ Mα contains a linearly dense subset of C(K). Lemma 19.22 says that {Mα}α<κ induces a PRI on C(K) onto subspaces of the form C(K/Mα) and, moreover, L/Mα separates the points of K/Mα for each α<κ. An argument as in the proof of Corollary 17.5 shows that C(K) has a one-to-one bounded τp-continuous linear operator into some c0(Γ ). In particular, L embeds into c0(Γ ). This completes the proof.

We now discuss stability of the class of Eberlein compact spaces. Straight from the deﬁnition, a closed subspace of an Eberlein compact space is Eberlein. It is not hard to see that a countable product of Eberlein compact spaces is Eberlein. A nontrivial result due to Benyamini, Rudin and Wage [49] says that the class of Eberlein compact spaces is stable under continuous images. It is interesting that the dual class of WCG Banach spaces is not stable under passing to subspaces (see [349]). Below we give a simple proof of the stability result mentioned in cases of totally disconnected spaces. It is based on the following lemma, which will be used later, too.

Lemma 19.23 Let K be a totally disconnected compact space. Then K is Eberlein if and only if Cp(K, 2) is σ -compact. U = U Proof Assume first that K is Eberlein, and let n∈ω n be a T0-separating family of clopen sets such that each Un is point-finite (Theorem 19.30). Observe that, identifying clopen sets with their characteristic functions, Un ∪{∅}is homeomorphic to the one-point compactification of a discrete space (∅ is the only possible accumulation point). Thus U is σ -compact. Finally, C(K,2) is, as a Boolean algebra, generated by (the characteristic functions of) U and therefore is σ -compact, too. = Now assume that Cp(K, 2) n∈ω Ln, where each Ln is compact. Let 1 L = L . n n n∈ω It is easy to check that L is compact in the pointwise topology. Finally, L clearly separates the points of K.

~~Theorem 19.32 Let f : K → L be a quotient map of totally disconnected compact spaces. If K is Eberlein, then so is L.~~

~~Proof It is clear that Cp(L, 2) embeds into Cp(K, 2) as a closed subspace. Apply Lemma 19.23.~~

We now discuss a very special class of compact spaces introduced by Tala- grand [388], which provides interesting examples concerning Eberlein compact spaces and their relatives. 462 19 Compact Spaces Generated by Retractions

We identify a Cantor cube 2S with the power set P(S). In this way, any family of sets becomes a totally disconnected Hausdorff topological space. A family of sets K is adequate if it satisﬁes the condition

~~x ∈ K ⇐⇒ [ x]<ω ⊆ K. (AD)~~

It is clear that adequate families are compact in the Cantor cube topology: The fact that x/∈ K is witnessed by a ﬁnite set that provides an appropriate neighborhood. An adequate compact space an adequate family considered as a (compact) topological space. Observe that every adequate compact space is Valdivia. It is rather clear that an adequate family is a Corson compact space if and only if it consists of countable sets (otherwise it contains a nonmetrizable Cantor cube, which is not Corson). Fix an adequate family K ⊆ P(S). We deﬁne the adequate dual of K to be the # topological space K = (S ∪{∞},τK ), where ∞ ∈/ S, all points of S are isolated in τK and a subbasic neighborhood of ∞ is of the form (S ∪{∞}) \ a, where a ∈ K. The next lemma shows the relation between K# and K.

~~Lemma 19.24 (Talagrand [388]) Let K be an adequate compact space. Then K# is homeomorphic to a closed subspace of Cp(K, 2) separating the points of K.~~

Proof Let K ⊆ P(S), and assume {α}∈K for every α ∈ S (otherwise replace S by K). Given α ∈ S,letα+ ={x ∈ K : α ∈ x}. We claim that K# is homeomorphic to X ={∅}∪{α+ : α ∈ S}, where elements of X are identified with their characteristic functions (and so X ⊆ Cp(K, 2)). We also need to show that X is closed. For the proof, observe that K is a compact semilattice when endowed with the binary operation (x, y) → x ∩ y.LetL be the set of all continuous homomorphisms of f : K → 2 satisfying f(∅) = 0. Here, 2 ={0, 1} is endowed with the usual semilattice structure: i ∧ j = min{i, j}. Identify elements of X (which are clopen subsets of K) with their characteristic functions. Observe that X ⊆ L, where ∅ corresponds to the zero homomorphism. It is clear that L is closed in Cp(K, 2) since if f ∈ C(K,2) is not a semilattice homomorphism, then this fact is witnessed by two elements of K that provide an appropriate neighborhood of f disjoint from L. Now fix φ ∈ L \ 0, and let a be the minimal element of φ−1(1). This element exists by compactness and by the fact that φ−1(1) is closed under finite meet (intersection). Notice that a must be finite since otherwise it would be an accumulation point of its finite subsets that are in φ−1(0).Nowφ is the unique f ∈ L satisfying the equations f(a)= 1 and (∀ b a) f(b) = 0. This shows that φ is isolated in L. Finally, observe that the characteristic function of α+ is an element of L, and ∅ corresponds to the zero homomorphism. In particular, a subbasic neighborhood of ∅ is of the form p− ={x ∈ L: p/∈ x}, where p ∈ K. 19.17 More on Eberlein compact spaces 463

In this way, we have shown that the map h: K# → X, deﬁned by h(α) = α+ and h(∞) =∅, is a homeomorphism. Finally, X is closed in Cp(K, 2) because L \ X consists of isolated points. It is obvious that X separates the points of K.

~~Below we characterize adequate Eberlein compact spaces.~~

Theorem 19.33 Let K be an adequate family of subsets of a set κ. Then K is = ∈ Eberlein compact if and only if κ n∈ω En, where for each n ω the family K ∩ P(En) consists of ﬁnite sets only.

# Proof By Lemmas 19.23 and 19.24, K is Eberlein if and only if K is σ -compact. It remains to observe that, assuming κ = K, a compact subset of K# is either ﬁnite or has the form A ∪{∞}, where A ⊆ κ and A ∩ x is ﬁnite for every x ∈ K. Indeed, if x ∈ K, then A ∩ x is a closed discrete subset of K#.

The result above gives a tool for ﬁnding non-Eberlein Corson compact spaces. Below we describe a speciﬁc subclass of adequate compact spaces coming from partially ordered sets. Given a poset P ,letK(P) denote the family of all chains (i.e., linearly ordered subsets) of P . Clearly, K(P) is adequate. The space K(P) is Corson if and only if all chains in P are countable. The next example was found by Alster and Pol [4]; the idea goes back to Sierpinski.«

Example 19.4 Let S ⊆ R be a set of cardinality ℵ1, and let be a fixed well order on S of order type ω1. Define s t if and only if both s ≤ t and s t hold, where ≤ denotes the usual linear order on S (taken from the reals). Then (S, ) is a poset in which all chains are countable since no uncountable subset of R can be well ordered. Thus the adequate family K(S) consisting of all chains in (S, ) is Corson compact. We claim that it is not Eberlein compact. = Suppose otherwise. By Theorem 19.33, S n∈ω Sn, where for each n all chains in Sn are finite. Let k ∈ ω be such that Sk is uncountable. Find C ⊆ Sk such that {x ∈ C : c

~~Theorem 19.34 (LeidermanÐSokolov [268]) Let T beatree. Then the space of chains K(T) is Eberlein compact if and only if T is special.~~

Proof If T is special, then the condition from Theorem 19.33 is satisfied. Assume = that K(T) is Eberlein, and apply Theorem 19.33. Thus, T n∈ω Tn, where all chains in Tn are finite. It suffices to show that each Tn is a countable union of an- tichains. However, Tn forms a tree in which all chains are finite. Recall that the ∈ { ∈ : } height of p Tn in Tn is the order type of the set x Tn x

~~Example 19.5 Denote by wQ the tree whose elements are well-ordered subsets of the rational numbers Q and s~~

~~Theorem 19.35 The tree wQ is not special.~~

Proof We show that the subtree σ Qconsisting of all bounded well-ordered subsets Q Q = of is not special. Suppose σ n∈ω An, where each An is an antichain. We inductively construct a sequence t0 ≤ t1 ≤ ... in σ Q and a sequence r0 ≥ r1 ≥ ... of rational numbers such that rn > sup tn and the following rule holds.

~~(*) If there exists s ∈ An with tn−1 ≤ s and sup s~~

Let us note that a result similar to Theorem 19.34 was obtained by Gruen- hage [193]: The space of all initial paths (i.e., downward closed chains) of a tree T is Eberlein if and only if the tree is special. As one can guess, nontrivial here is the “only if” part. Gruenhage’s result can be proved by arguments similar to those for adequate compact spaces, analyzing a slightly different “dual” space (analogous to the adequate dual). In this direction, perhaps the most “extreme” Corson compact space built on a certain tree was found by Todorceviˇ c.« We refer to [254, Example 5.5] for its short description and further references. It is also proved in [254] that this Corson compact κ space cannot be embedded into a cube R so that its range is dense in c0(κ). 19.17 More on Eberlein compact spaces 465

We have not discussed Talagrand and Gul’ko compact spaces here. We refer readers to Fabian’s book [147] for a nice presentation on these classes and corresponding WCD Banach spaces. One can prove that the examples described above are not Gul’ko compact. Examples of Gul’ko or Talagrand compact spaces that are not Eberlein can be found, for example, in [388] and [147, Chapter 8]. Chapter 20 Complementably Universal Banach Spaces

Abstract We deal with complementably universal Banach spaces. Assuming the continuum hypothesis, there exists a complementably universal Banach space of density ℵ1 for the class of Banach spaces with a projectional resolution of the identity. Similar methods produce a universal preimage for the class of Valdivia compacta of weight ℵ1.

~~20.1 Amalgamation lemma~~

A classical result of Pełczynski´ [327], says that there exists a separable Banach space E with a Schauder basis such that every other separable space with a basis is isomorphic to a complemented subspace of E (the space E is called complementably universal for the class of separable spaces with a basis). When looking for non-separable versions of this result, one needs ﬁrst to ask what kind of basis to consider. It turns out that, assuming the continuum hypothesis, the class of spaces with a countably norming Markushevich basis has a complementably universal object. We shall present its construction below. One has to mention a negative result due to Todorceviˇ c[´ 406]. For every Banach space E of size continuum and with Corson’s property (C), there exists a WLD Banach space F of size continuum that is not isomorphic to a subspace of any quotient of E. A topological counterpart, also proved [406], reads as follows: For every countably tight compact space K of weight continuum, there exists a Corson compact space L of weight continuum such that no closed subspace of K maps continuously onto L. A similar result on the nonexistence of universal Eberlein compact spaces was proved earlier by Argyros and Benyamini [12]. It is known that given two Banach spaces X and Y whose intersection Z = X ∩Y is closed in both spaces and for which both norms coincide on Z, there exists a Banach space W containing an isometric copy of X ∪ Y .Tobemoreprecise, there exist isometric embeddings i : X → W , j : Y → W such that i Z = j Z. Such a space W is called an amalgamation of X and Y . We need a special version of this fact, where Z is 1-complemented both in X and in Y .Thisisproved below.

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 467 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0_20, © Springer Science+Business Media, LLC 2011 468 20 Complementably Universal Banach Spaces

~~In the diagrams, we draw arrows to indicate linear isometric embeddings and for norm-one projections.~~

Lemma 20.1 Let X and Y be two Banach spaces with Z = X ∩ Y aclosedsub- space both in X and in Y . Assume further that P : X → X and Q: Y → Y are norm-one projections such that im P = Z = im Q. Then there exist a Banach space W , isometric embeddings i : X → W , j : Y → W and norm-one linear operators P : W → X, Q : W → Y such that P ◦ i = idX,Q◦ j = idY , and the following diagrams commute.

~~j j Q Q Y W Y W Y W Y W~~

~~⊆ i Q P ⊆ i Q P ~~

~~Z X Z X Z X ZX ⊆ ⊆ P P~~

Proof Let W = (X ⊕1 Y)/Δ, where ⊕1 denotes the direct sum with the 1 norm (i.e., (x, y) = x + y for x ∈ X, y ∈ Y ) and Δ ={(z, −z): z ∈ Z}. Recall that the quotient norm in W is given by the formula

(x, y) + Δ = inf ( x − z + x + z ). z∈Z Deﬁne i(x) = (x, 0) + Δ and j(y)= (0,y)+ Δ. Observe that i(x) ≤ x .On the other hand, x − z + z ≥ x , and therefore i(x) = x . Thus, i is an isometric embedding. By the same argument, j is an isometric embedding. Observe that if (x, y) − (x,y) ∈ Δ, then x = x − z, y = y + z,forsomez ∈ Z and hence

~~x + Qy = x − z + Q(y + z) = x − z + Qy + z = x + Qy.~~

~~Similarly, y + Px = y + Px. Thus, the following deﬁnitions are correct: P ((x, y) + Δ) = x + Qy and Q ((x, y) + Δ) = y + Px.~~

Clearly, P and Q are linear and im P ⊆ X + Z = X,imQ ⊆ Y + Z = Y . Fur- thermore, the inequality P ((x, y) + Δ) ≤ x + y holds whenever (x, y) + Δ = (x,y) + Δ. It follows that P ((x, y) + Δ) ≤ (x, y) + Δ .

~~Hence P ≤1. Similarly, Q ≤1. 20.2 Embedding-projection pairs 469~~

It is clear that P ◦i = idX and Q ◦j = idY . It is straightforward to check that the ﬁrst three diagrams from the statement of our lemma are commutative. It remains to check that P ◦ P = Q ◦ Q. For this, notice that im P = Z = im Q, and therefore QP x = Px and PQy= Qy for every x ∈ X, y ∈ Y . Thus,

~~PP((x, y) + Δ) = P(x+ Qy) = Px+ Qy = Q(P x + y) = QQ((x, y) + Δ).~~

~~This completes the proof.~~

~~Let us note that the ﬁrst part of the proof shows that W is an amalgamation of X and Y , without assuming that Z is 1-complemented in these spaces.~~

~~20.2 Embedding-projection pairs~~

In order to shorten some statements, we shall denote by ‡B the category of pairs of the form f = (i, P ), where i : X → Y , P : Y → X, are linear operators of norm ≤ 1 such that P ◦ i = idX and X, Y are separable Banach spaces. Given two compatible pairs f = (i, P ) and g = (j, Q) in ‡B, their composition is defined in the obvious way: f ◦ g = (j ◦ i, P ◦ Q). Observe that for every (i, P ) ∈ ‡B we have that i is a linear isometric embedding and i ◦ P : Y → Y is a projection of norm one (unless X ={0}). Conversely, given a norm-one projection Q: Y → Y with X = im Q, we get a ‡B-arrow (i, Q)˜ , where i is the inclusion X ⊆ Y and Q˜ is the corestriction of Q to its range (i.e., Q˜ : Y → X is defined by Qv˜ = Qv for v ∈ Y ). Given any category K, one can consider the category ‡K, defined in the same way as above. Our particular case is based on the category B of separable Banach spaces with linear operators of norm ≤ 1. In general, ‡K is called the category of embedding-projection pairs (see [134]). In the language of embedding-projection pairs, Lemma 20.1 says in particular that every pair of arrows f,g ∈ ‡B with the same domain can be amalgamated in ‡B. That is, there are f ,g ∈ ‡B such that f ◦ f = g ◦ g. Indeed, if f = (i, P ), g = (j, Q) are such that dom(i) = Z = dom(j) and X = dom(P ), Y = dom(Q),we may replace X and Y by their isometric copies so that X ∩ Y = Z and i, j become inclusions. It is important that Lemma 20.1 give a stronger version of amalgamation. This will be needed for universality. The following is the main technical lemma. ℵ 0 =ℵ = Lemma 20.2 Assume 2 1. There exists a Banach space U α<ω Uα such { } {1 } that Uα α<ω1 is a continuous chain of closed separable subspaces and Qα α<ω1 is a projectional resolution of the identity on U such that im Qα = Uα for α<ω1. Furthermore, the following conditions are satisfied.

~~(1)U0 ={0} (the trivial Banach space). 470 20 Complementably Universal Banach Spaces~~

(2) Given α<ω1 and a pair of norm-one linear operators i : Uα → Y , P : Y → Uα ◦ = such that Y is a separable Banach space and P i idUα , there exist β>α and norm-one linear operators j : Y → Uβ , R : Uβ → Y such that R ◦ j = idY and the following diagrams commute.

~~⊆ QαUβ Uα Uβ Uα Uβ~~

~~i j P R Y Y~~

Proof Let B be a fixed family of separable Banach spaces such that every separable Banach space is linearly isometric to exactly one element of B. It is well known and ℵ easy to compute that |B|=2 0 . For each separable Banach space X, we fix a linear isometry hX : X → Y such that Y ∈ B.Now,letF denote the family of all pairs f = (i, P ) such that X, Y ∈ B, i : X → Y , P : Y → X are linear operators of norm ≤ 1 and P ◦ i = idX. In other words, F consists of all arrows of ‡B whose domain ℵ and codomain are in B.Again,|F |=2 0 . ℵ 0 =ℵ × F { } Using the assumption 2 1, enumerate the set ω1 as (ξα,fα) α<ω1 so that for each (ξ, f ) ∈ ω1 × F there exist uncountably many α<ω1 satisfying (ξ, f ) = (ξα,fα). { } We shall construct inductively a continuous chain Uα α<ω1 of separable Banach { β } spaces and a sequence Qα α<β<ω1 of norm-one projections in such a way that the following conditions are satisfied: β (3) U0 ={0} and Qα : Uβ → Uα. =⇒ = η ◦ (4) ξ<η< Qξ Qξ Qη . = (5) If ξα <αand fα (iα,Pα) is such that dom(iα) is linearly isometric to Uξα , = ∈ = −1 ◦ then, setting h hUξα , there exists (j, R) ‡B such that dom(R) Uα, h P ◦ R = Qα and j ◦ i ◦ h is the inclusion U ⊆ U . α ξα α ξα α Condition (5) is visualized in the next two diagrams.

~~Qα ⊆ ξα Uξα Uα Uξα Uα~~

~~◦ −1 iα h j h ◦Pα R Y Y~~

Condition (3) tells us how to start the construction. Fix α<ω1, and suppose Uξ and η Qξ have been deﬁned for ξ<η<α.Ifα is a limit ordinal, let Uα be the completion α α = η of the union ξ<α Uξ and let Qξ be the unique projection satisfying Qξ Uη Qξ for every ξ<η<α(see the proof of Theorem 17.5). 20.3 A complementably universal Banach space 471

= + Suppose α ζ 1. If ξα <αor dom(iα) is not linearly isometric to Uξα ,weset = α = ζ : → Uα Uζ and Qξ Qα for ξ<α. So assume that ξα <αand h Uξα dom(iα) = = ◦ is the chosen linear isometry (i.e., h hUξα ). Let Y dom(Pα). Note that (iα −1 h, h ◦ Pα) ∈ ‡B. Using Lemma 20.1, we can ﬁnd W and (k, T ), (j, R) ∈ ‡B such that

~~− j ◦ i ◦ h = k U and h 1 ◦ P ◦ R = Qζ ◦ T. α α α ξα Actually, we are using a weaker form of Lemma 20.1, which gives the following two commutative diagrams.~~

~~Qζ ⊆ ξα Uξα Uζ Uξα Uζ~~

~~◦ −1 iα h k h ◦Pα T~~

~~Y W Y W j R~~

We may further assume that Uζ ⊆ W and k is the inclusion. We set Uα = W and α = ζ ◦ Qξ Qξ T for ξ<α. It is clear that conditions (4) and (5) are satisﬁed. The construction above gives a Banach space U = U (note that this α<ω1 α space is already complete). It follows from Theorem 17.5 (or rather from its proof) { β } { } that Qα α<β<ω1 induces a projectional resolution of the identity Qα α<ω1 on U β such that Qα Uβ = Qα for every α<β<ω1. It remains to check that condition (2) is satisﬁed. Fix (i, P ) ∈ ‡B such that i : Uα → Y for some α<ω1. Recall that there is a unique X ∈ B linearly isometric to Uα.Leth: Uα → X be the chosen isometry (i.e., = ∈ B = h hUα ). Without loss of generality, we may assume that Y . Notice that f (i ◦ h−1,h◦ P)is an element of F . Recall that we have used a “rich” enumeration of ω1 × F , so there exists β>αsuch that ξβ = α and fβ = f . Now condition (5) tells us that for some (j, R) ∈ ‡B we have that j ◦ (i ◦ h−1) ◦ h = j ◦ i is the −1 β inclusion Uα ⊆ Uβ and h ◦ (h ◦ P)◦ R = P ◦ R equals Qα . This shows (3) and completes the proof.

~~20.3 A complementably universal Banach space~~

~~We are now ready to prove the main result of this chapter.~~

~~ℵ Theorem 20.1 (Kubis[´ 249]) Assume the continuum hypothesis 2 0 =ℵ1. There exists a Banach space U with the following properties:~~

~~(a) dens U =ℵ1 and U has a projectional resolution of the identity. 472 20 Complementably Universal Banach Spaces~~

(b) Every Banach space with a projectional resolution of the identity and of density ≤ℵ1 is linearly isometric to a 1-complemented subspace of U. (c) Given two separable 1-complemented subspaces X, Y ⊆ U, for every bijective linear isometry h: X → Y there exists a bijective linear isometry h: U → U such that h X = h.

Proof We shall show that the space U from Lemma 20.2 has the desired properties. Property (a) was stated and proved already in Lemma 20.2. ≤ℵ { } = Fix a Banach space X of density 1 and with a PRI Pα α<ω1 .LetXα im Pα. = ={ } { } We may assume that P0 0(i.e.,X0 0 ). Note that Xα α<ω1 is a continuous chain of separable subspaces whose union is X.IfX is separable, we may assume β β that Xα = X for α>0. Let Qα = Qα Uβ and Pα = Pα Xβ . We shall construct inductively a sequence of pairs (jα,Rα) ∈ ‡B so that dom(jα) = Xα, dom(Rα) = Xδ(α) and for each ξ<ηthe following diagrams commute.

~~δ(η) ⊆ Qδ(ξ) ⊆ Uδ(ξ) Uδ(η) Uδ(ξ) Uδ(η) Uδ(ξ) Uδ(η)~~

~~jξ jη Rξ Rη Rξ Rη~~

~~Xξ Xη Xξ Xη Xξ Xη ⊆ η ⊆ Pξ~~

= ={ } = = = Since X0 U0 0 , we put δ(0) 0, and j0 idX0 R0.Fixβ>0, and assume that (jξ ,Rξ ) have been deﬁned for every ξ<β.Ifβ is a limit ordinal, we set = ∈ δ(β) supξ<β δ(ξ). There is a uniquely determined pair (jβ ,Rβ ) ‡B such that the diagrams above commute for every ξ<η≤ β. Suppose now that β = α + 1. Using Lemma 20.1, ﬁnd (k, S) ∈ ‡B and (, T ) ∈ ‡B with W = dom(S) = dom(T ), so that the following diagrams commute. k S k Uδ(α) W Uδ(α) W Uδ(α) W

~~jα Rα T Rα T~~

~~Xα Xα+1 Xα Xα+1 Xα Xα+1 ⊆ α+1 ⊆ Pα~~

~~We now use condition (2) of Lemma 20.2 to obtain δ>δ(α)and (m, A) ∈ ‡B,so that the following diagrams commute.~~

~~δ ⊆ Qδ(α) Uδ(α) Uδ Uδ(α) Uδ~~

~~k m S A W W 20.3 A complementably universal Banach space 473~~

Finally, we set δ(α+ 1) = δ and (jα+1,Rα+1) = (m ◦ , T ◦ A). It is straightforward to check that the condition above is satisﬁed. The construction above leads to an embedding j : X → U together with a norm- one operator R : U → X such that R ◦ j = idX. Thus, X is isometric to a 1- complemented subspace of U. This shows (b). Part (c) is proved by a standard back-and-forth argument. We sketch it below. Fix a bijective linear isometry h: X → Y between separable 1-complemented β β β β subspaces of U. First, let us denote by uα the ‡B-pair (ια ,Qα ), where ια : Uα → β Uβ is inclusion and Qα = Qα Uβ . We shall construct inductively ‡B-arrows fξ : Uα(ξ) → Uβ(ξ) and gξ : Uβ(ξ) → Uα(ξ+1) for ξ<ω1, so that

(i) fξ = (iξ ,Pξ ), gξ = (jξ ,Rξ ) and i0 extends h, ◦ = β(ξ) ◦ = β(ξ+1) (ii) gξ fξ uα(ξ) and fξ+1 gξ uβ(ξ) , and =⇒ β(η) ◦ = ◦ α(η) (iii) ξ<η uβ(ξ) fξ fη uα(ξ). ⊆ ⊆ ∈ Find α(0)<θ<ω1 such that X Uα0 and Y Uθ .Let(a, B) ‡B be such that a : X → Uθ is the composition of h with the inclusion Y ⊆ Uθ and B : Uθ → X is a fixed norm-one projection onto Y followed by h−1. Using Lemma 20.1, find (c, D) ∈ ‡B and (e, F ) ∈ ‡B so that the following diagrams commute, where S : Uα(0) → X is a fixed norm-one projection.

~~⊆ S ⊆ S X Uα(0) X Uα(0) X Uα(0) X Uα(0)~~

~~a e B F B B a e~~

~~Uθ W Uθ W Uθ W Uθ W c D c D~~

Using property (2) of Lemma 20.2, we ﬁnd β(0)>θ and (c,D) ∈ ‡B such that ◦ = β(0) ◦ = β(0) = ◦ ◦ c c ιθ and D D Qθ .Wesetf0 (c e,F D ). Using condition (2) of Lemma 20.2, ﬁnd g0 ∈ ‡B such that ◦ = α(1) α(1) g0 f0 ια(0)ια(0) .

Now fix 0 <δ<ω1, and suppose {fη}η<δ and {gη}η<δ have been defined. If δ is a limit ordinal, we define iδ and Pδ to be the unique map extending η<δ iη and η<δ Pη, respectively. This defines fδ. We define gδ in the same way and easily check that conditions (ii) and (iii) are satisfied. Suppose δ = ξ + 1. Using condition (2) of Lemma 20.2 twice, find fξ+1 and then gξ+1 such that ◦ = β(ξ+1) β(ξ+1) ◦ = α(ξ+2) α(ξ+2) fξ+1 gξ ιβ(ξ) ,Qβ(ξ) and gξ+1 fξ+1 ια(ξ+1),ια(ξ+1) . This finishes the construction. Finally, let h: U → U be such that h Uα(ξ) = ιβ(ξ) ◦ iα, where ια denotes the inclusion Uα ⊆ U. The existence of h is guaranteed by condition (ii). Furthermore, (i) says that h extends h. Finally, (iii) ensures that h is a bijective linear isometry. 474 20 Complementably Universal Banach Spaces

The construction above can be generalized to other categories. In particular, assuming the continuum hypothesis, there exists a Valdivia compact space K of density ℵ1 such that every other Valdivia compact space of the same density is homeomorphic to a retract of K. For a general, category-theoretic approach to this subject, we refer to [249]. In fact, the Banach space constructed in Lemma 20.2 is the so- called Fraïssé limit of the category ‡B. Originally, the notion of a Fraïssé limit belongs to model theory; it was developed by Fraïssé [167] and continued by Jóns- son [214]. The category-theoretic approach is due to Droste and Göbel [134] and was extended in Kubis[´ 249]. References

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A Compact line, 6, 367, 395 Adequate compact, 462 connectification, 397 Adequate dual, 462 jump, 396 Adequate family, 462 Compact resolution, 71 Almost quasi-Suslin, 221 Complementably universal space, 467 Amalgamation, 467 Complemented subspace, 355 Angelic, 76 Connectification, 397 Aronszajn continuum, 403 Continuous functor, 412 Aronszajn line, 403 Corson compact, 202, 426 Asplund space, 378 Countable Suslin number, 254 Averaging operator, 367 CS-barrelled, 57 CSPP, 268 B Cutting P-point, 416 Baire-like, 24 b-Baire-like space, 24 D Baire property, 15 D( · ), 375 Baire space, 14 Density condition, 170 Barrelled, 21 Diameter condition, 91 Biorthogonal system, 383 Distinguished space, 327 complete, 383 Docile, 321 Bornivorous, 24, 308 Dominating cardinal, 338 Bounded tightness, 302 Double arrow, 399 Bounding cardinal, 338 Dual metric, 239 Dyadic space, 232 C c0, 366 E C-quasibarrelled space, 277 ε-interchanges limits, 121 Canonical copy of c0, 366 Eberlein compact, 456 Cech-complete,ˇ 15 Elementary submodel/substructure, 356 Class RC , 411 Embedding-projection pairs, 469 Class R, 411 Extension operator, 402 Class G, 3, 243 Cofinal subset, 338 F Compact Family Corson, 426 T0-separating, 457 Eberlein, 456 First category, 13 Valdivia, 425 Fréchet–Urysohn space, 289

J. Kakol ˛ et al., Descriptive Topology in Selected Topics of Functional Analysis, 491 Developments in Mathematics 24, DOI 10.1007/978-1-4614-0529-0, © Springer Science+Business Media, LLC 2011 492 Index

G O (gDF )-space, 277 Operator γ -property, 309 extension, 402 G-basis, 306 regular, 401 Generating projections, 363 Gul’ko compact, 202 P π-base, 216 H P-point, 416 H(θ), 356 Plichko pair, 386 Hemicompact, 203 Plichko space, 386 Point I internal, 367 Increasing map, 395 Polish space, 16 Inductive limit space, 25 Poset, 369 Internal point, 367 σ -complete, 369 Interval topology, 395 antichain, 463 Inverse sequence, 406 closed, 369 continuous, 406 coﬁnal, 369 Inverse system, 405 directed, 369 bonding map, 405 special, 463 projection, 406 PRI, 372 retractive, 406 Primarily Lindelöf, 438 Projection, 355 K induced by (X, D, M), 363 K-analytic, 64 Projectional generator, 363 K-countably determined space, 64 Projectional resolution, 372 K-sequence, 296 Projectional sequence, 372 k-space, 174 Projectional skeleton, 369 kR -space, 203 commutative, 391 equivalence of, 375 L Projectively σ -compact, 207 (LB)-space, 25 Property (C), 279 (LF )-space, 25 Property (C1), 311 Lindelöf, 65 Property (C2), 311 Local π base at point, 216 Property (C3), 311 − Locally dense subspace, 59 Property (C3 ), 321 Property (C4), 318 M Pseudo-Lindelöf, 222 Mackey topology, 77 Pseudobase, 174 Map Pseudocompact, 33 increasing, 395 Markushevich basis, 383 Q countably norming, 383 Quasi-(LB)-space, 82 norming, 383 Quasi-Suslin, 65 Measurable function, 176 Quasibarrelled, 21 Monolithic set, 409 Monolithic space, 290 R Reﬂection principle, 356 N Regular (LM)-space, 308 Network, 68 Regular operator, 401 Norming space/set, 361 Regular SCP, 401 generating projections, 363 Relativization, 355 Nowhere dense, 13 Resolution, 1 Index 493

Retractional skeleton, 434 T commutative, 434 Talagrand compact, 117, 202 Three-space property, 193 S Tightness, 284 Σ-space, 386 Tightness m, 215 Σ-subset, 425 Topologically bounded, 33 σ -bounded, 207 Trans-separable, 157 σ -compact, 201 Transitive closure, 355 Σ-product, 380 Transitivity, 355 s-barrelled, 57 Tree, 463 Scattered, 229 special, 463 SCP, 355 regular, 401 U Separable complementation, 355 Ultrabornological space, 60 Separating Usco, 64 T0, 457 Sequence V projectional, 372 Valdivia compact, 425 Sequential, 237, 289 Space W complementably universal, 467 ω-cover, 227 norming, 361 Weak noncompactness, 130 nowhere separable, 397 Weakly compact density condition, 171 Split interval, 400 Weakly compactly generated Banach space, Strong condensation property, 271 261 Strongly WCG Banach space, 168 Weakly countably determined, 268 Strongly weakly K-analytic, 154 Weakly Lindelöf determined space, 440 Strongly web-bounding, 221 Web-bounded, 221 Strongly web-compact, 137 Web-bounding, 221 Superresolution, 310 Web-compact, 113 Suslin scheme, 91 WLD, 440